The rational numbers satisfy every algebraic law that the real numbers do. What they lack is limits: a sequence of rationals whose terms draw together need not converge to a rational. Analysis lives in \(\mathbb R\) because there every Cauchy sequence converges, and it is that property, far more than any formula, which lets one prove that solutions of equations exist.

A space of functions with a norm faces the same test. Take the continuous functions on \([0,1]\) , or the Riemann integrable ones, with the distance \(\int\lvert f-g\rvert\) . Both fail: there are sequences of step functions that are Cauchy for this distance and whose only possible limit is not Riemann integrable (Example 8.1). They are the rationals of the theory. The Lebesgue spaces are the reals. This is the real reason the Lebesgue integral displaced Riemann’s, more than its ability to integrate strange functions: \(L^p\) contains the limits of its own Cauchy sequences.

The chapter proves that, and then asks two companion questions. In what sense does an \(L^p\) -convergent sequence converge at individual points? And how can a general element of \(L^p\) , which is only an equivalence class of measurable functions, be approximated by functions one can actually handle?

8.1 Cauchy sequences converge

A sequence \((f_n)\) in \(L^p(\mu)\) is a Cauchy sequence if for every \(\varepsilon>0\) there is an \(N\) with \(\lVert f_m-f_n\rVert_p<\varepsilon\) for all \(m,n\ge N\) . The whole difficulty of proving convergence is that a limit has to be produced from nowhere. The key is a statement about series, in which the candidate limit is built pointwise.

All three convergence theorems of Chapters 4 and 5 are present: Minkowski bounds the partial sums, monotone convergence makes the bound survive the limit, and dominated convergence upgrades pointwise convergence to convergence in norm.

A complete normed vector space is called a Banach space (math.functional-analysis.banach-space), and for \(p=2\) , where the norm comes from the inner product \(\int fg\,d\mu\) , a Hilbert space. What can be done with that structure is the subject of Functional Analysis. The proof yields a second fact, as important in practice as the theorem.

8.2 Three kinds of convergence

Convergence in \(L^p\) says that an average of \(\lvert f_n-f\rvert^p\) tends to zero. Convergence almost everywhere is a statement about each point separately. Between them sits a third notion, which measures the size of the set where \(f_n\) and \(f\) differ noticeably.

No other implication holds in general. The four sequences below, three of them from Chapter 5, show this, and they are worth knowing by heart, since almost every question of the form “does this kind of convergence imply that kind” is settled by one of them.

Example 8.2: The typewriter sequence

For \(k\in\mathbb N\) write \(k=2^m+j\) with \(m\ge0\) and \(0\le j<2^m\) , and let

\[ f_k=\mathbf 1_{[\,j2^{-m},\,(j+1)2^{-m}\,]}\qquad\text{on } [0,1]. \]

So \(f_1\) is the indicator of \([0,1]\) ; \(f_2,f_3\) are the indicators of the two halves; \(f_4,\dots,f_7\) of the four quarters; and so on. Within each generation the interval moves from left to right like the carriage of a typewriter, and then returns with half the width.

Convergence in norm and in measure. For \(2^m\le k<2^{m+1}\) , \(\lVert f_k\rVert_p=2^{-m/p}\) for every finite \(p\) , and \(\lambda(\{f_k>\varepsilon\})=2^{-m}\) for \(0<\varepsilon<1\) . Both tend to \(0\) as \(k\to\infty\) . So \(f_k\to0\) in every \(L^p\) , \(p<\infty\) , and in measure.

No pointwise convergence anywhere. Fix \(x\in[0,1]\) . In every generation \(m\) the dyadic intervals of length \(2^{-m}\) cover \([0,1]\) , so at least one \(f_k\) with \(2^m\le k<2^{m+1}\) equals \(1\) at \(x\) ; and for \(m\ge2\) at least one equals \(0\) there, since the closed intervals of a generation containing \(x\) number at most two. (For \(m=1\) the point \(x=\tfrac12\) lies in both halves.) Hence the sequence \(f_k(x)\) takes the values \(0\) and \(1\) infinitely often and converges for no \(x\) at all. For \(x=\tfrac13\) , the indices \(k\le31\) with \(f_k(x)=1\) are exactly \(k=1,2,5,10,21\) , one in each generation.

The subsequence. Corollary 8.3 promises a subsequence converging a.e. Here \(f_{2^m}=\mathbf 1_{[0,2^{-m}]}\) will do: it tends to \(0\) at every \(x>0\) .

Figure 8.1 shows the first four generations and the points at which the vertical line \(x=\tfrac13\) is hit.

8.3 Dense classes of functions

An element of \(L^p\) is an abstract object. To prove something about all of them one usually proves it first for a convenient class and then extends by approximation. For this the class must be dense: every element of \(L^p\) must be a limit, in norm, of members of the class.

On the line the approximating functions can be made much more concrete. A step function is a finite linear combination of indicators of bounded intervals, and \(C_c(\mathbb R)\) denotes the continuous functions that vanish outside some bounded interval.

The proof is a chain of three approximations, each resting on an earlier chapter: an \(L^p\) function by a simple function (Chapter 3), a measurable set by a finite union of intervals (Chapter 2), and an interval by a trapezoid. The same statements hold on \(\mathbb R^d\) with rectangles in place of intervals.

Exercises

1. For the flattening block \(w_n=\tfrac1n\mathbf 1_{[0,n]}\) on \(\mathbb R\) compute \(\lVert w_n\rVert_p\) for \(1\le p\le\infty\) and decide, for each \(p\) , whether \(w_n\to0\) in \(L^p\) .

   Answer

   For \(p<\infty\) , \(\lVert w_n\rVert_p=(n^{-p}\cdot n)^{1/p}=n^{1/p-1}\) , and \(\lVert w_n\rVert_\infty=\tfrac1n\) . So \(w_n\to0\) in \(L^p\) for every \(p\in(1,\infty]\) , and not for \(p=1\) , where the norm is \(1\) for all \(n\) .

2. Write out the proof that \(L^\infty(\mu)\) is complete, stating exactly which null sets are discarded and why their union is null.

   Solution

   Let \((f_n)\) be a Cauchy sequence in \(L^\infty(\mu)\) ; fix one representative of each class.

   The null sets. For each pair \((m,n)\) let \(N_{m,n}=\{x:\lvert f_m(x)-f_n(x)\rvert>\lVert f_m-f_n\rVert_\infty\}\) . By the remark after Definition 7.2 the essential supremum is attained, so each \(N_{m,n}\) is null. There are countably many pairs, and a countable union of null sets is null by countable subadditivity; let \(N=\bigcup_{m,n}N_{m,n}\) .

   The pointwise limit. For \(x\notin N\) and all \(m,n\) , \(\lvert f_m(x)-f_n(x)\rvert\le\lVert f_m-f_n\rVert_\infty\) . The right side is small for large \(m,n\) , so \((f_n(x))\) is a Cauchy sequence of real numbers and has a limit \(f(x)\) . Put \(f=0\) on \(N\) . Then \(f=\lim_nf_n\mathbf 1_{N^c}\) is measurable.

   Convergence in norm. Given \(\varepsilon>0\) choose \(n_0\) with \(\lVert f_m-f_n\rVert_\infty<\varepsilon\) for \(m,n\ge n_0\) . For \(x\notin N\) and \(n\ge n_0\) , letting \(m\to\infty\) in \(\lvert f_m(x)-f_n(x)\rvert<\varepsilon\) gives \(\lvert f(x)-f_n(x)\rvert\le\varepsilon\) . So \(\lvert f-f_n\rvert\le\varepsilon\) off the null set \(N\) , that is, \(\lVert f-f_n\rVert_\infty\le\varepsilon\) for \(n\ge n_0\) . In particular \(\lVert f\rVert_\infty\le\lVert f_{n_0}\rVert_\infty+\varepsilon<\infty\) , so \(f\in L^\infty\) . The point of collecting all the exceptional sets into one null set at the start is that afterwards the argument is simply uniform convergence on \(N^c\) .

3. Let \(f_n\in L^1(\mu)\) with \(\lVert f_{n+1}-f_n\rVert_1\le2^{-n}\) for all \(n\) . Prove that \(f_n\) converges almost everywhere and in \(L^1\) , and explain why the typewriter sequence does not contradict this.

   Solution

   Put \(g_n=f_{n+1}-f_n\) . Then \(\sum_n\lVert g_n\rVert_1\le\sum_n2^{-n}<\infty\) , so Lemma 8.1 with \(p=1\) applies: the series \(\sum_ng_n\) converges absolutely a.e. and in \(L^1\) . Its partial sums are \(\sum_{n<N}g_n=f_N-f_1\) . Hence \(f_N\) converges a.e., and in \(L^1\) , to \(f=f_1+\sum_ng_n\in L^1\) . This is the first half of the proof of Theorem 8.2 with the subsequence already given.

   The typewriter sequence converges to \(0\) in \(L^1\) and at no point, and it is Cauchy in \(L^1\) . There is no contradiction, because it is not rapidly Cauchy: within generation \(m\) consecutive terms are indicators of adjacent intervals of length \(2^{-m}\) , so \(\lVert f_{k+1}-f_k\rVert_1=2\cdot2^{-m}\) for each of the \(2^m-1\) consecutive pairs in that generation, and those differences alone sum to \(2-2^{1-m}\) . Summing over generations, \(\sum_k\lVert f_{k+1}-f_k\rVert_1=\infty\) . A Cauchy sequence need not have summable increments; it always has a subsequence that does, and the subsequence \(f_{2^m}\) is one.

4. For the typewriter sequence of Example 8.2 and the point \(x=\tfrac15\) , list the indices \(k\le63\) with \(f_k(x)=1\) .

   Answer

   \(k=1,2,4,9,19,38\) . In generation \(m\) the point \(\tfrac15\) lies in the dyadic interval with \(j=\lfloor2^m/5\rfloor\) , giving \(j=0,0,0,1,3,6\) for \(m=0,\dots,5\) and \(k=2^m+j\) .

5. Prove that limits in measure are unique up to null sets: if \(f_n\to f\) and \(f_n\to g\) in measure, then \(f=g\) almost everywhere.

   Hint

   For \(\varepsilon>0\) , \(\{\lvert f-g\rvert>2\varepsilon\}\subseteq\{\lvert f-f_n\rvert>\varepsilon\}\cup\{\lvert f_n-g\rvert>\varepsilon\}\) for every \(n\) . Let \(n\to\infty\) to get \(\mu(\{\lvert f-g\rvert>2\varepsilon\})=0\) , then take the union over \(\varepsilon=\tfrac1m\) .

6. Let \(1\le p<\infty\) , let \(f_n\to f\) in measure, and suppose \(\lvert f_n\rvert\le g\) for all \(n\) with \(g\in L^p\) . Prove that \(f_n\to f\) in \(L^p\) .

   Solution

   Suppose, for contradiction, that \(\lVert f_n-f\rVert_p\) does not tend to \(0\) . Then there are \(\delta>0\) and a subsequence \((f_{n_j})\) with \(\lVert f_{n_j}-f\rVert_p\ge\delta\) for all \(j\) . This subsequence still converges to \(f\) in measure, so by Proposition 8.5(b) it has a further subsequence \((f_{n_{j_i}})\) converging to \(f\) almost everywhere.

   Along that sub-subsequence, \(\lvert f_{n_{j_i}}\rvert\le g\) , and passing to the limit, \(\lvert f\rvert\le g\) a.e. Hence

   \[ \lvert f_{n_{j_i}}-f\rvert^p\le(2g)^p, \]

   which is integrable because \(g\in L^p\) , and the left side tends to \(0\) a.e. By dominated convergence \(\lVert f_{n_{j_i}}-f\rVert_p\to0\) , contradicting \(\lVert f_{n_j}-f\rVert_p\ge\delta\) .

   The pattern is worth isolating. Dominated convergence needs convergence almost everywhere, which convergence in measure supplies only along subsequences. But a sequence of real numbers tends to \(0\) as soon as every subsequence has a further subsequence tending to \(0\) , and that principle transfers the theorem from subsequences to the whole sequence. So the dominated convergence theorem remains true with “in measure” in place of “almost everywhere”.

7. A student argues: “\(f_n\to f\) in \(L^1\) , so \(f_n(x)\to f(x)\) for almost every \(x\) , so by Fatou’s lemma \(\int\lvert f\rvert\le\liminf\int\lvert f_n\rvert\) .” The conclusion is true. Identify the false step and repair the argument.

   Answer

   The false step is “so \(f_n(x)\to f(x)\) for almost every \(x\) ”; the typewriter sequence converges in \(L^1\) and nowhere pointwise. Repair: choose a subsequence along which \(\int\lvert f_{n}\rvert\) tends to \(\liminf_n\int\lvert f_n\rvert\) , then by Corollary 8.3 a further subsequence converging to \(f\) a.e., and apply Fatou’s lemma to that. (Here the conclusion also follows at once from \(\lvert\lVert f_n\rVert_1-\lVert f\rVert_1\rvert\le\lVert f_n-f\rVert_1\) , with equality of the limits.)

8. Among the functions that are constant on \([0,\tfrac12)\) and on \([\tfrac12,1]\) , find the one closest to \(x^2\) in \(L^2(0,1)\) , and compute the distance.

   Solution

   Let \(s=c_1\mathbf 1_{[0,1/2)}+c_2\mathbf 1_{[1/2,1]}\) . Then

   \[ \lVert x^2-s\rVert_2^2=\int_0^{1/2}(x^2-c_1)^2\,dx+\int_{1/2}^1(x^2-c_2)^2\,dx, \]

   and the two constants can be chosen independently. For an interval \(I\) of length \(\ell\) , the quadratic \(c\mapsto\int_I(x^2-c)^2\,dx\) is minimized where its derivative \(-2\int_I(x^2-c)\,dx\) vanishes, that is, at the average \(c=\frac1\ell\int_Ix^2\,dx\) . So

   \[ c_1=2\int_0^{1/2}x^2\,dx=\frac1{12},\qquad c_2=2\int_{1/2}^1x^2\,dx=2\Bigl(\frac13-\frac1{24}\Bigr)=\frac7{12}. \]

   At the minimizer, \(\int_I(x^2-c)^2=\int_Ix^4-\ell c^2\) , so

   \[ \lVert x^2-s\rVert_2^2=\int_0^1x^4\,dx-\frac12\Bigl(\frac1{144}+\frac{49}{144}\Bigr)=\frac15-\frac{25}{144}=\frac{19}{720}, \]

   and the distance is \(\sqrt{19/720}\approx0.1624\) . The best step function takes on each piece the mean value of the target there. This is the orthogonal projection of \(x^2\) onto the two-dimensional space of such step functions, a fact about \(L^2\) that the other \(L^p\) spaces do not share.

9. Show that \(C_c(\mathbb R)\) is not dense in \(L^\infty(\mathbb R)\) by exhibiting a bounded function whose distance from every element of \(C_c(\mathbb R)\) is at least \(1\) .

   Hint

   Take the constant function \(1\) . Any \(g\in C_c(\mathbb R)\) vanishes outside a bounded interval, and the complement of that interval has positive measure, so \(\lvert1-g\rvert=1\) on a set of positive measure.

10. For \(f(x)=e^{-\lvert x\rvert}\) on \(\mathbb R\) and \(h>0\) compute \(\lVert\tau_hf-f\rVert_1\) in closed form, and verify its limits as \(h\to0\) and as \(h\to\infty\) .

    Solution

    For \(h>0\) the function \(\tau_hf(x)=e^{-\lvert x-h\rvert}\) is the reflection of \(f\) in the point \(x=\tfrac h2\) , so \(\lvert\tau_hf-f\rvert\) is symmetric about \(\tfrac h2\) , and \(\tau_hf\ge f\) exactly for \(x\ge\tfrac h2\) . Hence

    \[ \lVert\tau_hf-f\rVert_1=2\int_{h/2}^\infty\bigl(e^{-\lvert x-h\rvert}-e^{-x}\bigr)\,dx . \]

    Compute the two pieces. First, \(\int_{h/2}^\infty e^{-x}\,dx=e^{-h/2}\) . Second,

    \[ \int_{h/2}^\infty e^{-\lvert x-h\rvert}\,dx=\int_{h/2}^he^{x-h}\,dx+\int_h^\infty e^{-(x-h)}\,dx=(1-e^{-h/2})+1 . \]

    Therefore

    \[ \lVert\tau_hf-f\rVert_1=2\bigl(2-e^{-h/2}-e^{-h/2}\bigr)=4\bigl(1-e^{-h/2}\bigr). \]

    As \(h\to0\) this behaves like \(2h\) and tends to \(0\) , as Corollary 8.8 requires. As \(h\to\infty\) it tends to \(4=2\lVert f\rVert_1\) : the two copies of \(f\) become effectively disjoint, and the norm of the difference is the sum of the norms. For example, at \(h=1\) the value is \(4(1-e^{-1/2})\approx1.5739\) .

11. Let \(f\in L^1(\mathbb R)\) . Prove that \(\int_{\mathbb R}f(x)\cos(tx)\,dx\to0\) as \(t\to\infty\) , by treating indicators of intervals first and then using Theorem 8.7.

    Solution

    Indicators. For a bounded interval with endpoints \(a<b\) and \(t>0\) ,

    \[ \Bigl\lvert\int_a^b\cos(tx)\,dx\Bigr\rvert=\Bigl\lvert\frac{\sin(tb)-\sin(ta)}{t}\Bigr\rvert\le\frac2t\longrightarrow0 . \]

    Step functions. A step function \(s\) is a finite linear combination of such indicators, so \(\int s(x)\cos(tx)\,dx\to0\) by linearity.

    General \(f\) . Let \(\varepsilon>0\) . By Theorem 8.7 there is a step function \(s\) with \(\lVert f-s\rVert_1<\varepsilon\) . Since \(\lvert\cos\rvert\le1\) ,

    \[ \Bigl\lvert\int f(x)\cos(tx)\,dx\Bigr\rvert\le\int\lvert f-s\rvert\,dx+\Bigl\lvert\int s(x)\cos(tx)\,dx\Bigr\rvert<\varepsilon+\Bigl\lvert\int s(x)\cos(tx)\,dx\Bigr\rvert . \]

    The last term is less than \(\varepsilon\) for all large \(t\) . Hence the upper limit of the left side is at most \(2\varepsilon\) , for every \(\varepsilon\) .

    This is the Riemann–Lebesgue lemma, and the proof is the density method in its purest form: an explicit computation on the simplest functions, and one inequality, uniform in \(t\) , that lets the statement pass to limits in \(L^1\) . Dominated convergence would not have worked here, because \(f(x)\cos(tx)\) does not converge pointwise as \(t\to\infty\) .

12. Prove that for \(1\le p<\infty\) the step functions with rational values and rational endpoints form a countable dense subset of \(L^p(\mathbb R)\) , and that \(L^\infty(\mathbb R)\) has no countable dense subset.

    Hint

    Rational step functions form a countable set; approximate real coefficients and endpoints by rational ones and use Theorem 8.7. For \(L^\infty\) , the functions \(\mathbf 1_{[0,t]}\) , \(t>0\) , are uncountably many and pairwise at distance \(1\) ; a dense set must contain a point within \(\tfrac12\) of each, and these points are distinct.

13. Suppose \(f_n\to f\) in \(L^p\) and \(f_n\to g\) almost everywhere. Prove that \(f=g\) almost everywhere.

    Answer

    By Corollary 8.3 a subsequence \(f_{n_k}\to f\) a.e. The same subsequence converges to \(g\) a.e. Limits of real sequences are unique, so \(f=g\) outside the union of two null sets.

14. Use Theorem 8.9 to prove that every Lebesgue measurable \(f:[a,b]\to\mathbb R\) is the almost-everywhere limit of a sequence of continuous functions.

    Hint

    For each \(n\) take compact \(K_n\) with \(\lambda([a,b]\setminus K_n)<2^{-n}\) and \(f\) continuous on \(K_n\) ; extend \(f\) from \(K_n\) to a continuous \(g_n\) on \([a,b]\) by linear interpolation across the gaps. Then \(g_n=f\) on \(K_n\) , and by Proposition 1.12 almost every \(x\) lies in \(K_n\) for all large \(n\) .

15. Let \(f_n\) be the continuous function on \([0,1]\) that is \(0\) on \([0,\tfrac12]\) , \(1\) on \([\tfrac12+\tfrac1n,1]\) , and linear between. Compute \(\lVert f_n-\mathbf 1_{(1/2,1]}\rVert_1\) and conclude that the continuous functions with the \(L^1\) norm are not complete.

    Solution

    The difference \(\mathbf 1_{(1/2,1]}-f_n\) vanishes outside \((\tfrac12,\tfrac12+\tfrac1n)\) , and on that interval it decreases linearly from \(1\) to \(0\) . Its graph is a triangle with base \(\tfrac1n\) and height \(1\) , so

    \[ \lVert f_n-\mathbf 1_{(1/2,1]}\rVert_1=\frac1{2n}\longrightarrow0 . \]

    Hence \((f_n)\) converges in \(L^1(0,1)\) , so it is a Cauchy sequence there, and each \(f_n\) is continuous. Suppose some continuous \(g\) on \([0,1]\) had \(\lVert f_n-g\rVert_1\to0\) . Limits in \(L^1\) are unique, so \(g=\mathbf 1_{(1/2,1]}\) almost everywhere. Then \(g=0\) a.e. on \([0,\tfrac12]\) and \(g=1\) a.e. on \((\tfrac12,1]\) ; by continuity \(g=0\) on all of \([0,\tfrac12]\) and \(g=1\) on all of \([\tfrac12,1]\) , which is impossible at the point \(\tfrac12\) .

    So the space of continuous functions on \([0,1]\) with the norm \(\lVert\cdot\rVert_1\) contains a Cauchy sequence with no limit in the space: it is not complete. By Theorem 8.7 it is dense in \(L^1(0,1)\) , and by Theorem 8.2 that space is complete. In this precise sense \(L^1\) is to the continuous functions what \(\mathbb R\) is to \(\mathbb Q\) .

16. Let \(\mu(X)<\infty\) . Prove that \(f_n\to f\) in measure if and only if every subsequence of \((f_n)\) has a further subsequence converging to \(f\) almost everywhere.

    Hint

    One direction is Proposition 8.5(b) applied to subsequences. For the other, argue by contradiction as in Exercise 8.6: if \(\mu(\{\lvert f_n-f\rvert>\varepsilon\})\) does not tend to \(0\) , extract a subsequence along which it stays above some \(\delta\) , then a further subsequence converging a.e., and apply Proposition 8.5(c).

17. Let \(1<p<\infty\) with conjugate \(q\) , \(f\in L^p(\mathbb R)\) , and \(g\in L^q(\mathbb R)\) , both Borel. Prove that \(f*g\) is defined everywhere, bounded by \(\lVert f\rVert_p\lVert g\rVert_q\) , and uniformly continuous.

    Solution

    Defined and bounded. Fix \(x\) . The function \(y\mapsto f(x-y)\) is Borel, and by invariance of Lebesgue measure under translation and reflection its \(L^p\) norm is \(\lVert f\rVert_p\) . By Hölder’s inequality \(y\mapsto f(x-y)g(y)\) is integrable, with

    \[ \lvert(f*g)(x)\rvert\le\int\lvert f(x-y)\rvert\,\lvert g(y)\rvert\,dy\le\lVert f\rVert_p\,\lVert g\rVert_q . \]

    So \(f*g\) is defined at every point, not merely almost everywhere, and is bounded.

    Uniformly continuous. Let \(\tilde f(u)=f(-u)\) , so that \(f(x-y)=\tilde f(y-x)=(\tau_x\tilde f)(y)\) . For any \(x\) and \(h\) , Hölder’s inequality gives

    \[ \lvert(f*g)(x+h)-(f*g)(x)\rvert\le\int\lvert\tau_{x+h}\tilde f-\tau_x\tilde f\rvert\,\lvert g\rvert\,dy\le\lVert\tau_{x+h}\tilde f-\tau_x\tilde f\rVert_p\,\lVert g\rVert_q=\lVert\tau_h\tilde f-\tilde f\rVert_p\,\lVert g\rVert_q, \]

    the last step by translation invariance. The right side does not depend on \(x\) , and it tends to \(0\) as \(h\to0\) by Corollary 8.8, since \(\tilde f\in L^p\) with \(p<\infty\) . That is uniform continuity. The hypothesis \(p<\infty\) was used only here; if \(p=\infty\) and \(q=1\) , apply the same argument with the roles of \(f\) and \(g\) exchanged.