The Lebesgue Integral

The Lebesgue integral In this second part of the course the basic theory of the Lebesgue integral is presented. Here I follow an idea of Jan Mikusiński, of completing the space of step functions on the line under the L1 norm but in such a way that the limiting objects are seen directly as functions (defined almost everywhere). There are other places you can find this, for instance the book of Debnaith and Mikusiński [1]. Here I start from the Riemann integral, since this is a prerequisite of the course; this streamlines things a little. The objective is to arrive at a working knowledge of Lebesgue integration as quickly as seems acceptable, to pass on to the discussion of Hilbert space and then to more analytic questions. So, the treatment of the Lebesgue integral here is intentionally compressed, while emphasizing the completeness of the spaces L1 and L2 . In lectures everything is done for the real line but in such a way that the extension to higher dimensions – carried out partly in the text but mostly in the problems – is not much harder. 1. Integrable functions Recall that the Riemann integral is defined for a certain class of bounded functions u : [a, b] −→ C (namely the Riemann integrable functions) which includes all continuous functions. It depends on the compactness of the interval and the boundedness of the function, but can be extended to an ‘improper integral’ on the whole real line for which however some of the good properties fail. This is NOT what we will do. Rather we consider the space of continuous functions ‘with compact support’: (2.1) Cc (R) = {u : R −→ C; u is continuous and ∃ R such that u(x) = 0 if |x| > R}. Thus each element u ∈ Cc (R) vanishes outside an interval [−R, R] where the R depends on the u. Note that the support of a continuous function is defined to be the complement of the largest open set on which it vanishes (or as the closure of the set of points at which it is non-zero – make sure you see why these are the same). Thus (2.1) says that the support, which is necessarily closed, is contained in some interval [−R, R], which is equivalent to saying it is compact. Lemma 2.1. The Riemann integral defines a continuous linear functional on Cc (R) equipped with the L1 norm Z Z u = lim u(x)dx, R→∞ [−R,R] R Z 1 = lim kuk |u(x)|dx, (2.2) L R→∞ [−R,R] Z | u| ≤ kukL1 . R The limits here are trivial in the sense that the functions involved are constant for large R. Proof. These are basic properties of the Riemann integral see Rudin [4].  Note that Cc (R) is a normed space with respect to kukL1 as defined above; that it is not complete is one of the main reasons for passing to the Lebesgue integral. With this small preamble we can directly define the ‘space’ of Lebesgue integrable functions on R. Definition 2.1. A function f : R −→ C is Lebesgue integrable, written f ∈ n P L (R), if there exists a series with partial sums fn = wj , wj ∈ Cc (R) which is 1 j=1 absolutely summable, XZ (2.3) |wj | < ∞ j and such that (2.4) X |wj (x)| < ∞ =⇒ lim fn (x) = n→∞ j X wj (x) = f (x). j This is a somewhat convoluted definition which you should think about a bit. Its virtue is that it is all there. The problem is that it takes a bit of unravelling. Before we go any further note that the sequence wj obviously determines the sequence of partial sums fn , both in Cc (R) but the converse is also true since w1 = f1 , wk = fk − fk−1 , k > 1, Z X XZ (2.5) |wj | < ∞ ⇐⇒ |fk − fk−1 | < ∞. j k>1 You might also notice that can we do some finite manipulation, for instance replace the sequence wj by X (2.6) W1 = wj , Wk = wN +k−1 , k > 1 j≤N and nothing much changes, since the convergence conditions in (2.3) and (2.4) are properties only of the tail of the sequences and the sum in (2.4) for wj (x) converges if and only if the corresponding sum for Wk (x) converges and then converges to the same limit. Before massaging the definition a little, let me give a simple example and check that this definition does include continuous functions defined on an interval and extended to be zero outside – the theory we develop will include the usual Riemann integral although I will not quite prove this in full, but only because it is not particularly interesting. Lemma 2.2. If f ∈ C([a, b]) then ( f (x) ˜ (2.7) f (x) = 0 if x ∈ [a, b] otherwise is an integrable function. Proof. Just ‘add legs’ to f˜ by considering the sequence  0 if x < a − 1/n or x > b + 1/n,    (1 + n(x − a))f (a) if a − 1/n ≤ x < a, (2.8) fn (x) =   (1 − n(x − b))f (b) if b < x ≤ b + 1/n,  f (x) if x ∈ [a, b]. This is a continuous function on each of the open subintervals in the description with common limits at the endpoints, so fn ∈ Cc (R). By construction, fn (x) → f˜(x) for each x ∈ R. Define the sequence wj which has partial sums the fn , as in (2.5) above. Then wj = 0 in [a, b] for j > 1 and it can be written in terms of the ‘legs’ ( 0 if x < a − 1/n, x ≥ a ln = (1 + n(x − a)) if a − 1/n ≤ x < a, ( 0 if x ≤ b, x > b + 1/n rn = (1 − n(x − b)) if b ≤ x ≤ b + 1/n, as |wn (x)| = (ln − ln−1 )|f (a)| + (rn − rn−1 )|f (b)|, n > 1. (2.9) It follows that Z |wn (x)| = (|f (a)| + |f (b)|) n(n − 1) so {wn } is an absolutely summable sequence showing that f˜ ∈ L1 (R).  Returning to the definition, notice that we only say ‘there exists’ an absolutely summable sequence and that it is required to converge to the function only at points at which the pointwise sequence is absolutely summable. At other points anything is permitted. So it is not immediately clear that there are any functions not P satisfying this condition. Indeed if there was a sequence like wj above with |wj (x)| = ∞ always, then (2.4) would represent no restriction at all. So the j point of the definition is that absolute summability – a condition on the integrals in (2.3) – does imply something about (absolute) convergence of the pointwise series. Let us reenforce this idea with another definition:Definition 2.2. A set E ⊂ R is said to be of measure zero in the sense of Lebesgue (which is pretty much always the meaning here) if there is a series n P PR gn = vj , vj ∈ Cc (R) which is absolutely summable, |vj | < ∞, and such that j=1 (2.10) j X j |vj (x)| = ∞ ∀ x ∈ E. Notice that we do not require E to be precisely the set of points at which the series in (2.10) diverges, only that it does so at all points of E, so E is just a subset of the set on which some absolutely summable series of functions in Cc (R) does not converge absolutely. So any subset of a set of measure zero is automatically of measure zero. To introduce the little trickery we use to unwind the definition above, consider first the following (important) result. Lemma 2.3. Any finite union of sets of measure zero is a set of measure zero. Proof. Since we can proceed in steps, it suffices to show that the union of two sets of measure zero has measure zero. So, let the two sets be E and F and two corresponding absolutely summable sequences, as in Definition 2.2, be vj and wj . Consider the alternating sequence ( vj if k = 2j − 1 is odd (2.11) uk = wj if k = 2j is even. Thus {uk } simply interlaces the two sequences. It follows that uk is absolutely summable, since X X X (2.12) kuk kL1 = kvj kL1 + kwj kL1 . j k j P Moreover, the pointwise series |uk (x)| diverges precisely where one or other of P P k the two series |vj (x)| or |wj (x)| diverges. In particular it must diverge on j j E ∪ F which is therefore, from the definition, a set of measure zero.  The definition of f ∈ L1 (R) above certainly requires that the equality on the right in (2.4) should hold outside a set of measure zero, but in fact a specific one, the one on which the series on the left diverges. Using the same idea as in the lemma above we can get rid of this restriction. n P Proposition 2.1. If f : R −→ C and there exists a series fn = wj with j=1 P wj ∈ Cc (R) which is absolutely summable, so kwj kL1 < ∞, and a set E ⊂ R of j measure zero such that (2.13) x ∈ R \ E =⇒ f (x) = lim fn (x) = n→∞ ∞ X wj (x) j=1 then f ∈ L1 (R). Recall that when one writes down an equality such as on the right in (2.13) one is ∞ P implicitly saying that wj (x) converges and the equality holds for the limit. We j=1 will call a sequence as the wj above an ‘approximating series’ for f ∈ L1 (R). 1 This is indeed a refinement of Pthe definition since all f ∈ L (R) arise this way, taking E to be the set where |wj (x)| = ∞ for a series as in the defintion. j Proof. By definition of a set of measure zero there is some series vj as in (2.10). Now, consider the series obtained by alternating the terms between wj , vj and −vj . Explicitly, set   wk uj = vk   −vk (x) (2.14) if j = 3k − 2 if j = 3k − 1 if j = 3k. This defines a series in Cc (R) which is absolutely summable, with X X X kvk kL1 . kwk kL1 + 2 (2.15) kuj (x)kL1 = j k k The same sort of identity is true for the pointwise series which shows that X X X (2.16) |uj (x)| < ∞ iff |wk (x)| < ∞ and |vk (x)| < ∞. j k k So if the pointwise series on the left converges absolutely, then x ∈ / E, by definition and hence, using (2.13), we find that X X (2.17) |uj (x)| < ∞ =⇒ f (x) = uj (x) j j since the sequence of partial sumsPof the uj cycles through fn , fn (x) + vn (x), then fn (x) and then to fn+1 (x). Since |vk (x)| < ∞ the sequence |vn (x)| → 0 so (2.17) k indeed follows from (2.13).  This is the trick at the heart of the definition of integrability above. Namely we can manipulate the series involved in this sort of way to prove things about the elements of L1 (R). One point to note is that if wj is an absolutely summable series in Cc (R) then P  |wj (x)| when this is finite j (2.18) F (x) = =⇒ F ∈ L1 (R). 0 otherwise The sort of property (2.13), where some condition holds on the complement of a set of measure zero is so commonly encountered in integration theory that we give it a simpler name. Definition 2.3. A condition that holds on R \ E for some set of measure zero, E, is said to hold almost everywhere. In particular we write (2.19) f = g a.e. if f (x) = g(x) ∀ x ∈ R \ E, E of measure zero. Of course as yet we are living dangerously because we have done nothing to show that sets of measure zero are ‘small’ let alone ‘ignorable’ as this definition seems to imply. Beware of the trap of ‘proof by declaration’ ! Now Proposition 2.1 can be paraphrased as ‘A function f : R −→ C is Lebesgue integrable if and only if it is the pointwise sum a.e. of an absolutely summable series in Cc (R).’ 2. Linearity of L1 The word ‘space’ is quoted in the definition of L1 (R) above, because it is not immediately obvious that L1 (R) is a linear space, even more importantly it is far from obvious that the integral of a function in L1 (R) is well defined (which is the point of the exercise after all). In fact we wish to define the integral to be Z XZ (2.20) f= wn R n where wn ∈ Cc (R) is any ‘approximating series’ meaning now as the wj in Propsition 2.1. This is fine in so far as the series P R on the right (of complex numbers) does converge – since we demanded that |wn | < ∞ so this series converges n absolutely – but not fine in so far as the answer might well depend on which series we choose which ‘approximates f ’ in the sense of the definition or Proposition 2.1. So, the immediate aim is to prove these two things. First we will do a little more than prove the linearity of L1 (R). Recall that a function is ‘positive’ if it takes only non-negative values. Proposition 2.2. The space L1 (R) is linear (over C) and if f ∈ L1 (R) the real and imaginary parts, Re f, Im f are Lebesgue integrable as are their positive parts and as is also the absolute value, |f |. For a real Lebesgue integrable function there is an approximating sequence as in Proposition 2.1 which is real and if f ≥ 0 the sequence of partial sums can be arranged to be non-negative. Proof. We first consider the real part of a function f ∈ L1 (R). Suppose wn ∈ Cc (R) is an approximating series as Rin Proposition 2.1. Then consider vn = Re wn . R This is absolutely summable, since |vn | ≤ |wn | and X X (2.21) wn (x) = f (x) =⇒ vn (x) = Re f (x). n n Since the left identity holds a.e., so does the right and hence Re f ∈ L1 (R) by Proposition 2.1. The same argument with the imaginary parts shows that Im f ∈ L1 (R). This also shows that a real element has a real approximating sequence. The fact that the sum of two integrable functions is integrable really is a simple consequence of Proposition 2.1 and Lemma 2.3. Indeed, if f, g ∈ L1 (R) have approximating series wn and vn as in Proposition 2.1 then un = wn +vn is absolutely summable, XZ XZ XZ (2.22) |un | ≤ |wn | + |vn | n n n and X n wn (x) = f (x), X n vn (x) = g(x) =⇒ X un (x) = f (x) + g(x). n The first two conditions hold outside (probably different) sets of measure zero, E and F, so the conclusion holds outside E ∪ F which is of measure zero. Thus f + g ∈ L1 (R). The case of cf for c ∈ C is more obvious. The proof that |f | ∈ L1 (R) if f ∈ L1 (R) is similar but perhaps a little trickier. Again, let {wn } be an approximating series as in the definition showing that f ∈ L1 (R). To make a series for |f | we can try the ‘obvious’ thing. Namely we know that n X X (2.23) wj (x) → f (x) if |wj (x)| < ∞ j=1 j so certainly it follows that | n X wj (x)| → |f (x)| if X j=1 |wj (x)| < ∞. j So, set (2.24) v1 (x) = |w1 (x)|, vk (x) = | k X wj (x)| − | j=1 k−1 X wj (x)| ∀ x ∈ R. j=1 Then, for sure, (2.25) N X k=1 vk (x) = | N X wj (x)| → |f (x)| if j=1 X |wj (x)| < ∞. j So equality holds off a set of measure zero and we only need to check that {vj } is an absolutely summable series. The triangle inequality in the ‘reverse’ form ||v| − |w|| ≤ |v − w| shows that, for k > 1, (2.26) k X |vk (x)| = || wj (x)| − | j=1 k−1 X wj (x)|| ≤ |wk (x)|. j=1 Thus (2.27) XZ k |vk | ≤ XZ |wk | < ∞ k so the vk ’s do indeed form an absolutely summable series and (2.25) holds almost everywhere, so |f | ∈ L1 (R). For a positive function this last argument yields a real approximating sequence with positive partial sums.  By combining these results we can see again that if f, g ∈ L1 (R) are both real valued then (2.28) f+ = max(f, 0), max(f, g), min(f, g) ∈ L1 (R). Indeed, the positive part, f+ = 21 (|f | + f ), max(f, g) = g + (f − g)+ , min(f, g) = − max(−f, −g). 3. The integral on L1 Next we want to show that the integral is well defined via (2.20) for any approximating series. From Propostion 2.2 it is enough to consider only real functions. For this, recall a result concerning a case where uniform convergence of continuous functions follows from pointwise convergence, namely when the convergence is monotone, the limit is continuous, and the space is compact. It works on a general compact metric space but we can concentrate on the case at hand. Lemma 2.4. If un ∈ Cc (R) is a decreasing sequence of non-negative functions such that limn→∞ un (x) = 0 for each x ∈ R then un → 0 uniformly on R and Z un = 0. (2.29) lim n→∞ Proof. Since all the un (x) ≥ 0 and they are decreasing (which really means not increasing of course) if u1 (x) vanishes at x then all the other un (x) vanish there too. Thus there is one R > 0 such that un (x) = 0 if |x| > R for all n, namely one that works for u1 . So we only need consider what happens on [−R, R] which is compact. For any  > 0 look at the sets Sn = {x ∈ [−R, R]; un (x) ≥ }. This can also be written Sn = u−1 n ([, ∞)) ∩ [−R, R] and since un is continuous it follows that Sn is closed and hence compact. Moreover the fact that the un (x) are decreasing means that Sn+1 ⊂ Sn for all n. Finally, \ Sn = ∅ n since, by assumption, un (x) → 0 for each x. Now the property of compact sets in a metric space that we use is that if such a sequence of decreasing compact sets has empty intersection then the sets themselves are empty from some n onwards. This means that there exists N such that supx un (x) <  for all n > N. Since  > 0 was arbitrary, un → 0 uniformly. One of the basic properties of the Riemann integral is that the integral of the limit of a uniformly convergent sequence (even of Riemann integrable functions but here continuous) is the limit of the sequence of integrals, which is (2.29) in this case.  We can easily extend this in a useful way – the direction of monotonicity is reversed really just to mentally distinquish this from the preceding lemma. Lemma 2.5. If vn ∈ Cc (R) is any increasing sequence such that limn→∞ vn (x) ≥ 0 for each x ∈ R (where the possibility vn (x) → ∞ is included) then Z (2.30) lim vn dx ≥ 0 including possibly + ∞. n→∞ Proof. This is really a corollary of the preceding lemma. Consider the sequence of functions ( 0 if vn (x) ≥ 0 (2.31) wn (x) = −vn (x) if vn (x) < 0. Since this is the maximum of two continuous functions, namely −vn and 0, it is continuous and it vanishes for large x, so wn ∈ Cc (R). Since vn (x) is increasing, wn is decreasing and it follows that lim wn (x) = 0 for all x – either it gets there for some finite n and then stays 0 or the limit of vn (x) is zero. Thus Lemma 2.4 applies to wn so Z lim wn (x)dx = 0. R R Now, vn (x) ≥ −wn (x) for all x, so for each Rn, vn ≥ − wn . From properties of the Riemann integral, vn+1 ≥ vn implies that vn dx is an increasing sequence and it is bounded below by one that converges to 0, so (2.30) is the only possibility.  n→∞ R From this result applied carefully we see that the integral behaves sensibly for absolutely summable series. Lemma 2.6. P RSuppose un ∈ Cc (R) is an absolutely summable series of real-valued functions, so |un |dx < ∞, and also suppose that n X (2.32) un (x) = 0 a.e. n then XZ (2.33) un dx = 0. n R Proof. RAs already noted, the series (2.33) does converge, since the inequality | un dx| ≤ |un |dx shows that it is absolutely convergent (hence Cauchy, hence convergent). If E is a set of measure zero such that (2.32) holds on the complement then we can modify un as in (2.14) by adding and subtracting a non-negative absolutely summable sequence vk which diverges absolutely on E. For the new sequence un (2.32) is strengthened to X X (2.34) |un (x)| < ∞ =⇒ un (x) = 0 n n and the conclusion (2.33) holds for the new sequence if and only if it holds for the old one. Now, we need to get ourselves into a position to apply Lemma 2.5. To do this, just choose some integer N (large but it doesn’t matter yet) and consider the sequence of functions – it depends on N but I will suppress this dependence – (2.35) U1 (x) = N +1 X un (x), Uj (x) = |uN +j (x)|, j ≥ 2. n=1 This P R is a sequence in Cc (R) and it is absolutely summable – the convergence of |Uj |dx only depends on the ‘tail’ which is the same as for un . For the same j reason, X (2.36) |Uj (x)| < ∞ ⇐⇒ X |un (x)| < ∞. n j Now the sequence of partial sums (2.37) gp (x) = p X Uj (x) = N +1 X un (x) + n=1 j=1 p X |uN +j | j=2 is increasing with p – since we are adding non-negative functions. If the two equivalent conditions in (2.36) hold then (2.38) X n un (x) = 0 =⇒ N +1 X un (x) + n=1 ∞ X |uN +j (x)| ≥ 0 =⇒ lim gp (x) ≥ 0, j=2 p→∞ since we are only increasing each term. On the other hand if these conditions do not hold then the tail, any tail, sums to infinity so (2.39) lim gp (x) = ∞. p→∞ Thus the conditions of Lemma 2.5 hold for gp and hence N +1 Z X Z X |uj (x)|dx ≥ 0. (2.40) un + n=1 j≥N +2 Using the same inequality as before this implies that ∞ Z X X Z (2.41) |uj (x)|dx. un ≥ −2 n=1 j≥N +2 This is true for any N and as N → ∞, limN →∞ P R |uj (x)|dx = 0. So j≥N +2 the fixed number on the left in (2.41), which is what we are interested in, must be non-negative. In fact the signs in the argument can be reversed, considering instead (2.42) h1 (x) = − N +1 X un (x), hp (x) = |uN +p (x)|, p ≥ 2 n=1 and the final conclusion is the opposite inequality in (2.41). That is, we conclude what we wanted to show, that ∞ Z X un = 0. (2.43) n=1  Finally then we are in a position to show that the integral of an element of L1 (R) is well-defined. Proposition 2.3. If f ∈ L1 (R) then Z XZ un (2.44) f = lim n→∞ n is independent of the approximating sequence, un , used to define it. Moreover, Z X Z N uk |, | |f | = lim N →∞ Z (2.45) | f| ≤ Z lim n→∞ k=1 Z |f − |f | and n X uj | = 0. j=1 So in some sense the definition of the Lebesgue integral ‘involves no cancellations’. There are various extensions of the integral which do exploit cancellations – I invite you to look into the definition of the Henstock integral (and its relatives). R Proof. The uniqueness of f follows from Lemma 2.6. Namely, if un and u0n are two series approximating f as in Proposition 2.1 then the real and imaginary parts of the difference u0n − un satisfy the hypothesis of Lemma 2.6 so it follows that XZ XZ un = u0n . n n Then the first part of (2.45) follows from this definition of the integral applied to |f | and the approximating series for |f | devised in the proof of Proposition 2.2. The inequality XZ XZ |un |, un | ≤ (2.46) | n n which follows from the finite inequalities for the Riemann integrals XZ XZ XZ |un | |un | ≤ un | ≤ | n≤N n≤N n gives the second part. The final part follows by applying the same arguments to the series {uk }k>n , n P as an absolutely summable series approximating f − uj and observing that the j=1 integral is bounded by Z ∞ Z n X X |uk | → 0 as n → ∞. uk | ≤ (2.47) |f − k=n+1 k=1  4. Summable series in L1 (R) The next thing we want to know is when the ‘norm’, which is in fact only a R seminorm, on L1 (R), vanishes. That is, when does |f | = 0? One way is fairly easy. The full result we are after is:R Proposition 2.4. For an integrable function f ∈ L1 (R), the vanishing of |f | implies that f is a null function in the sense that f (x) = 0 ∀ x ∈ R \ E where E is of measure zero. R Conversely, if (2.48) holds then f ∈ L1 (R) and |f | = 0. (2.48) R Proof. The main part of this is the first part, that the vanishing of |f | implies that f is null. The converse is the easier direction in the sense that we have already done it. Namely, if f is null in the sense of (2.48) then |f | is the limit a.e. of the absolutely summable series with all Rterms 0. It follows from the definition of the integral above that |f | ∈ L1 (R) and |f | = 0.  For the forward argument we will use the following more technical result, which is also closely related to the completeness of L1 (R) (note the small notational difference, L1 is the Banach space which is the quotient by the null functions, see below). Proposition 2.5. If fn ∈ L1 (R) is an absolutely summable series, meaning PR that |fn | < ∞, then n (2.49) E = {x ∈ R; X |fn (x)| = ∞} has measure zero. n If f : R −→ C satisfies (2.50) f (x) = X n fn (x) a.e. then f ∈ L1 (R), Z f= XZ fn , n Z (2.51) | Z Z f| ≤ |f | = lim n→∞ | Z lim n→∞ |f − n X j=1 n X fj | ≤ XZ |fj | and j fj | = 0. j=1 This basically says we can replace ‘continuous function of compact support’ by ‘Lebesgue integrable function’ in the definition and get the same result. Of course this makes no sense without the original definition, so what we are showing is that iterating it makes no difference – we do not get a bigger space. Proof. The proof is very like the proof of completeness via absolutely summable series for a normed space outlined in the preceding chapter. 1 P R By assumption each fn ∈ L (R), so there exists a sequence un,j ∈ Cc (R) with |un,j | < ∞ and j (2.52) X |un,j (x)| < ∞ =⇒ fn (x) = j X un,j (x). j We might hope that f (x) is given by the sum of the un,j (x) over both n and j, but in general, this double series is not absolutely summable. However we can replace it by one that is. For each n choose Nn so that X Z (2.53) |un,j | < 2−n . j>Nn This is possible by the assumed absolute summability – the tail of the series therefore being small. Having done this, we replace the series un,j by X (2.54) u0n,1 = un,j (x), u0n,j (x) = un,Nn +j−1 (x) ∀ j ≥ 2, j≤Nn summing the first Nn terms. This still sums to fn on the same set as in (2.52). So in fact we can simply replace un,j by u0n,j and we have in addition the estimate Z XZ (2.55) |u0n,j | ≤ |fn | + 2−n+1 ∀ n. j This follows from the triangle inequality since, using (2.53), Z Z Z N X XZ (2.56) |u0n,1 + u0n,j | ≥ |u0n,1 | − |u0n,j | ≥ |u0n,1 | − 2−n j=2 j≥2 R and the left side converges to |fn | by (2.45) as N → ∞. Using (2.53) again gives (2.55). Dropping the primes from the notation and denoting the new series again as un,j we can let vk be some enumeration of the un,j – using the standard diagonalization procedure for instance. This gives a new series of continuous functions of compact support which is absolutely summable since N Z XZ XZ X |un,j | ≤ ( |fn | + 2−n+1 ) < ∞. (2.57) |vk | ≤ k=1 n n,j Using the freedom to rearrange absolutely convergent series we see that X X XX X (2.58) |un,j (x)| < ∞ =⇒ f (x) = vk (x) = un,j (x) = fn (x). n,j k n j n The set where (2.58) fails is a set of measure zero, by definition. Thus f ∈ L1 (R) and (2.49) also follows. To get the final result (2.51), rearrange the double series for the integral (which is also absolutely convergent).  For the moment we only need the weakest part, (2.49), of this. To paraphrase this, for any absolutely summable series of integrable functions the absolute pointwise series converges off a set of measure zero – it can only diverge on a set of measure zero. It is rather shocking but thisR allows us to prove the rest of Proposition 2.4! Namely, suppose f ∈ L1 (R) and |f | = 0. Then Proposition 2.5 applies to the series with each term being |f |. This is absolutely summable since all the integrals are zero. So it must converge pointwise except on a set of measure zero. Clearly it diverges whenever f (x) 6= 0, Z (2.59) |f | = 0 =⇒ {x; f (x) 6= 0} has measure zero which is what we wanted to show to finally complete the proof of Proposition 2.4. 5. The space L1 (R) At this point we are able to define the standard Lebesgue space (2.60) L1 (R) = L1 (R)/N , N = {null functions} and R to check that it is a Banach space with the norm (arising from, to be pedantic) |f |. Theorem 2.1. The quotient space L1 (R) defined by (2.60) is a Banach space in which the continuous functions of compact support form a dense subspace. The elements of L1 (R) are equivalence classes of functions (2.61) [f ] = f + N , f ∈ L1 (R). That is, we ‘identify’ two elements of L1 (R) if (and only if) their difference is null, which is to say they are equal off a set of measure zero. Note that the set which is ignored here is not fixed, but can depend on the functions. Proof. For an element of L1 (R) the integral of the absolute value is welldefined by Propositions 2.2 and 2.4 Z 1 (2.62) k[f ]kL = |f |, f ∈ [f ] and gives a semi-norm on L1 (R). It follows from Proposition 1.5 that on the quotient, k[f ]k is indeed a norm. The completeness of L1 (R) is a direct consequence of Proposition 2.5. Namely, to show a normed space is complete it is enough to check that any absolutely summable series converges. If [fj ] is an absolutely summable series in L1 (R) then fj is absolutely summable in L1 (R) and by Proposition 2.5 the sum of the series exists so we can use (2.50) to define f off the set E and take it to be zero on E. Then, f ∈ L1 (R) and the last part of (2.51) means precisely that Z X X |f − fj | = 0 (2.63) lim k[f ] − [fj ]kL1 = lim n→∞ n→∞ j 1, is absolutely summable so un − φn → 0 almost everywhere, and hence φn → f off a set of measure zero.  So replacing ‘simple functions’ by continuous functions in Definition 2.8 makes no difference – and the same for approximation by elements of L1 (R). The measurable functions form a linear space since if f and g are measurable and fn , gn are sequences of simple functions as required by the definition then c1 fn (x) + c2 f2 (x) → c1 f (x) + c2 g(x) on the intersection of the sets where fn (x) → f (x) and gn (x) → g(x) which is the complement of a set of measure zero. Now, from the discussion above, we know that each element of L1 (R) is measurable. Conversely: Lemma 2.11. A function f : R −→ C is an element of L1 (R) if and only if it is measurable and there exists F ∈ L1 (R) such that |f | ≤ F almost everywhere. Proof. If f is measurable there exists a sequence of simple functions fn such that fn → f almost everywhere. The real part, Re f, is also measurable as the limit almost everywhere of Re fn and from the hypothesis | Re f | ≤ F. We know that there exists a sequence of simple functions gn , gn → F almost everywhere and 0 ≤ gn ≤ G for another element G ∈ L1 (R). Then set   if Re fn (x) > gn (x) gn (x) (2.106) un (x) = Re fn (x) if − gn (x) ≤ Re fn (x) ≤ gn (x)   −gn (x) if Re fn (x) < −gn (x). Thus un = max(vn , −gn ) where vn = min(Re fn , gn ) so un is simple and un → f almost everywhere. Since |un | ≤ G it follows from Lebesgue Dominated Convergence that Re f ∈ L1 (R). The same argument shows Im f = − Re(if ) ∈ L1 (R) so f ∈ L1 (R) as claimed.  11. The spaces Lp (R) We use Lemma 2.11 as a model: Definition 2.9. For 1 ≤ p < ∞ we set Lp (R) = {f : R −→ C; f is measurable and |f |p ∈ L1 (R)}. (2.107) For p = ∞ we set L∞ (R) = {f : R −→ C; f measurable and ∃ C s.t. |f (x)| ≤ C a.e} (2.108) Observe that, in view of Lemma 2.10, the case p = 2 gives the same space as Definition 2.4. Proposition 2.9. For each 1 ≤ p < ∞, Z  p1 p |u| (2.109) kukLp = is a seminorm on the linear space Lp (R) vanishing only on the null functions and making the quotient Lp (R) = Lp (R) N into a Banach space. Proof. The real part of an element of Lp (R) is in Lp (R) since it is measurable and | Re f |p ≤ |f |p so | Re f |p ∈ L1 (R). Similarly, Lp (R) is linear; it is clear that cf ∈ Lp (R) if f ∈ Lp (R) and c ∈ C and the sum of two elements, f, g, is measurable and satisfies |f + g|p ≤ 2p (|f |p + |g|p ) so |f + g|p ∈ L1 (R). We next strengthen (2.107) to the approximation condition that there exists a sequence of simple functions vn such that vn → f a.e. and |vn |p ≤ F ∈ L1 (R) a.e. (2.110) which certainly implies (2.107). As in the proof of Lemma 2.11, suppose f ∈ Lp (R) is real and choose fn real-valued simple functions and converging to f almost everywhere. Since |f |p ∈ L1 (R) there is a sequence of simple functions 0 ≤ hn such that |hn | ≤ F for some F ∈ L1 (R) and hn → |f |p almost everywhere. Then set 1 gn = hnp which is also a sequence of simple functions and define vn by (2.106). It follows that (2.110) holds for the real part of f but combining sequences for real and imaginary parts such a sequence exists in general. The advantage of the approximation condition (2.110) is that it allows us to conclude that the triangle inequality holds for kukLp defined by (2.109) since we know it for simple functions and from (2.110) it follows that |vn |p → |f |p in L1 (R) so kvn kLp → kf kLp . Then if wn is a similar sequence for g ∈ Lp (R) (2.111) kf +gkLp ≤ lim sup kvn +wn kLp ≤ lim sup kvn kLp +lim sup kwn kLp = kf kLp +kgkLp . n n n The other two conditions being clear it follows that kukLp is a seminorm on Lp (R). The vanishing of kukLp implies that |u|p and hence u ∈ N and the converse follows immediately. Thus Lp (R) = Lp (R) N is a normed space and it only remains to check completeness. We know that completeness is equivalent to the convergence of any absolutely summable series. So, we can suppose fn ∈ Lp (R) have  p1 X Z (2.112) |fn |p < ∞. n Consider the sequence gn = fn χ[−R,R] for some fixed R > 0. This is in L1 (R) and (2.113) 1 kgn kL1 ≤ (2R) q kfn kLp by the integral form of Hölder’s inequality (2.114) Z 1 1 f ∈ Lp (R), g ∈ Lq (R), + = 1 =⇒ f g ∈ L1 (R) and | f g| ≤ kf kLp |kgkLq p q which can be proved by the same approximation argument as above, see Problem ??. Thus the series gn is absolutely summable inPL1 and so converges absolutely almost everywhere. It follows that the series fn (x) converges absolutely almost n P everywhere – since it is just gn (x) on [−R, R], to a function, f. n R So, we only need show that f ∈ Lp (R) and that |f − Fn |p → 0 as n → ∞ n n P P where Fn = fk . By Minkowski’s inequality we know that hn = ( |fk |)p has k=1 k=1 bounded L1 norm, since 1 p k|hn |kL1 = k (2.115) n X k=1 |fk |kLp . ≤ X kfk kLp . k Thus, hn is an increasing sequence of functions in L1 (R) with bounded integral, so by the Monotonicity Lemma it converges a.e. to a function h ∈ L1 (R). Since |Fn |p ≤ h and |Fn |p → |f |p a.e. it follows by Dominated convergence that 1 X kfn kLp (2.116) |f |p ∈ L1 (R), k|f |p kLp 1 ≤ n p and hence f ∈ L (R). Applying the same reasoning to f − Fn which is the sum of the series starting at term n + 1 gives the norm convergence: X (2.117) kf − Fn kLp ≤ kfk kLp → 0 as n → ∞. k>n  A function f : R −→ C is locally integrable if ( f (x) x ∈ [−N, N ] (2.118) F[−N,N ] = =⇒ F[−N,N ] ∈ L1 (R) ∀ N. 0 x if |x| > N So any continuous function on R is locally integrable as is any element of L1 (R). Lemma 2.12. The locally integrable functions form a linear space, L1loc (R) and Lp (R) = {f ∈ L1loc (R); |f |p ∈ L1 (R)} 1 ≤ p < ∞ (2.119) L∞ (R) = {f ∈ L1loc (R); sup |f (x)| < ∞ for some E of measure zero.} R\E The proof is left as an exercise. 12. Lebesgue measure In case anyone is interested in how to define Lebesgue measure from where we are now we can just use the integral. Definition 2.10. A set A ⊂ R is measurable if its characteristic function χA is locally integrable. A measurable set A has finite measure if χA ∈ L1 (R) and then Z (2.120) µ(A) = χA is the Lebesgue measure of A. If A is measurable but not of finite measure then µ(A) = ∞ by definition. We know immediately that any interval (a, b) is measurable (indeed whether open, semi-open or closed) and has finite measure if and only if it is bounded – then the measure is b − a. Proposition 2.10. The complement of a measurable set is measurable and any countable union or countable intersection of measurable sets is measurable. Proof. The first part follows from the fact that the constant function 1 is locally integrable and hence χR\A = 1 − χA is locally integrable if and only if χA is locally integrable. Notice the relationship between a characteristic function and the set it defines:(2.121) χA∪B = max(χA , χB ), χA∩B = min(χA , χB ). S If we have a sequence of sets An then Bn = k≤n Ak is clearly an increasing sequence of sets and X (2.122) χBn → χB , B = An n is an increasing sequence which converges pointwise (at each point it jumps to 1 somewhere and then stays or else stays at 0.) Now, if we multiply by χ[−N,N ] then (2.123) fn = χ[−N,N ] χBn → χB∩[−N,N ] is an increasing sequence of integrable functions – assuming that is that the Ak ’s are measurable – with integral bounded above, by 2N. Thus by the S monotonicity lemma the limit is integrable so χB is locally integrable and hence n An is measurable. For countable intersections the argument is similar, with the sequence of characteristic functions decreasing.  Corollary 2.1. The (Lebesgue) measurable subsets of R form a collection, M, of the power set of R, including ∅ and R which is closed under complements, countable unions and countable intersections. Such a collection of subsets of a set X is called a ‘σ-algebra’ – so a σ-algebra Σ in a set X is a collection of subsets of X containing X, ∅, the complement of any element and countable unions and intersections of any element. A (positive) measure is usually defined as a map µ : Σ −→ [0, ∞] with µ(∅) = 0 and such that [ X (2.124) µ( En ) = µ(En ) n n for any sequence {Em } of sets in Σ which are disjoint (in pairs). As for Lebesgue measure a set A ∈ Σ is ‘measurable’ and if µ(A) is not of finite measure it is said to have infinite measure – for instance R is of infinite measure in this sense. Since the measure of a set is always non-negative (or undefined if it isn’t measurable) this does not cause any problems and in fact Lebesgue measure is countably additive as in (2.124) provided we allow ∞ as a value of the measure. It is a good exercise to prove this! 13. Higher dimensions I have never actually covered this in lectures – there is simply not enough time. Still it is worth knowing that the Lebesgue integral in higher dimensional Euclidean spaces can be obtained following the same line of reasoning. So, we want – with the advantage of a little more experience – to go back to the beginning and define L1 (Rn ), L1 (Rn ), L2 (Rn ) and L2 (Rn ). In fact relatively little changes but there are some things that one needs to check a little carefully. The first hurdle is that I am not assuming that you have covered the Riemann integral in higher dimensions; it is in my view a rather pointless thing to do anyway. Fortunately we do not really need that since we can just iterate the one-dimensional Riemann integral for continuous functions. So, define (2.125) Cc (Rn ) = {u : Rn −→ C; continuous and such that u(x) = 0 for |x| > R} where of course the R can depend on the element. Now, if we hold say the last n − 1 variables fixed, we get a continuous function of one variable which vanishes when |x| > R : u(·, x2 , . . . , xn ) ∈ Cc (R) for each (x2 , . . . , xn ) ∈ Rn−1 . (2.126) So we can integrate it and get a function Z (2.127) I1 (x2 , . . . , xn ) = u(x, x1 , . . . , xn ), I1 : Rn−1 −→ C. R Lemma 2.13. For each u ∈ Cc (Rn ), I1 ∈ Cc (Rn−1 ). Proof. Certainly if |(x2 , . . . , xn )| > R then u(·, x2 , . . . , xn ) ≡ 0 as a function of the first variable and hence I1 = 0 in |(x2 , . . . , xn )| > R. The continuity follows from the uniform continuity of a function on the compact set |x| ≤ R, |(x2 , . . . , xn ) ≤ R of Rn . Thus given  > 0 there exists δ > 0 such that |x − x0 | < δ, |y − y 0 |Rn−1 < δ =⇒ |u(x, y) − u(x0 , y 0 )| < . (2.128) From the standard estimate for the Riemann integral, Z R 0 (2.129) |I1 (y) − I1 (y )| ≤ |u(x, y) − u(x, y 0 )|dx ≤ 2R −R 0 if |y − y | < δ. This implies the (uniform) continuity of I1 . Thus I1 ∈ Cc (Rn−2 )  The upshot of this lemma is that we can integrate again, and hence a total of n times and so define the (iterated) Riemann integral as Z Z R Z R Z R (2.130) u(z)dz = ··· u(x1 , x2 , x3 , . . . , xn )dx1 dx2 . . . dxn ∈ C. Rn −R −R −R Lemma 2.14. The interated Riemann integral is a well-defined linear map (2.131) Cc (Rn ) −→ C which satisfies Z Z (2.132) | u| ≤ |u| ≤ (2R)n sup |u| if u ∈ Cc (Rn ) and u(z) = 0 in |z| > R. Proof. This follows from the standard estimate in one dimension.  Now, one slightly annoying thing is that we would really want to know that the integral is independent of the order of integration. In fact it is not hard – see Problem XX. Again using properties of the one-dimensional Riemann integral we find:Lemma 2.15. The iterated integral Z |u| kukL1 = (2.133) Rn is a norm on Cc (Rn ). Definition 2.11. The space L1 (Rn ) is defined to consist of those functions f : Rn −→ C such that there exists a sequence {fn } which is absolutely summable with respect to the L1 norm and such that X X |fn (x)| < ∞ =⇒ fn (x) = f (x). (2.134) n n Now you can go through the whole discusion above in this higher dimensional case, and the only changes are really notational! Things get a littlem more complicated in the discussion of change of variable. This is covered in the problems. There are also a few other theorems it is good to know!