The Strategy Behind the Proof
The completeness theorem says that every Cauchy sequence in \(L^p(X,\mu)\) converges to an element of that same space. The central difficulty is not selecting a candidate limit at the start: a general Cauchy sequence may have no obvious pointwise limit. Instead, we first select a subsequence whose successive differences have summable \(L^p\) norms. Those differences can then be added pointwise, outside a null set, to construct a limit.
There are two distinct conclusions to justify. The pointwise construction must produce a measurable function in \(L^p\), not merely a measurable function. It must also give convergence in the \(L^p\) norm. Once a subsequence converges in norm, the result from “Completeness of \(L^p\)” that a convergent subsequence determines a Cauchy sequence gives convergence of the original sequence.
The proof below treats \(1<p<\infty\). The completeness results for \(L^1\), \(L^2\), and \(L^\infty\) were established earlier in this course. The summable-difference argument itself also works for \(p=1\), but the endpoint theorem need not be proved again here.
A Summable-Difference Criterion
We begin with a criterion that turns control of successive differences into convergence. The initial \(L^p\) hypothesis is essential: controlling differences alone cannot ensure that the limit belongs to \(L^p\).
Proof. Define the nonnegative measurable functions $$ S_N=\sum_{k=1}^{N}|v_{k+1}-v_k|. $$ By Minkowski’s inequality for finite sums, $$ \|S_N\|_p\leq\sum_{k=1}^{N}\|v_{k+1}-v_k\|_p \leq A, $$ where \(A=\sum_{k=1}^{\infty}\|v_{k+1}-v_k\|_p<\infty\). The sequence \((S_N^p)\) increases pointwise to \(S^p\), where $$ S=\sum_{k=1}^{\infty}|v_{k+1}-v_k| $$ is allowed initially to take the value \(+\infty\). The Monotone Convergence Theorem gives $$ \int_X S^p\,d\mu =\lim_{N\to\infty}\int_X S_N^p\,d\mu =\lim_{N\to\infty}\|S_N\|_p^p \leq A^p. $$ Thus \(S\in L^p(X,\mu)\), so \(S\) is finite almost everywhere.
Let \(G=\{x\in X:S(x)<\infty\}\). This is a measurable set, and its complement has measure zero. For every \(x\in G\), the series of absolute successive differences converges. For integers \(m>n\), $$ |v_m(x)-v_n(x)| \leq\sum_{k=n}^{m-1}|v_{k+1}(x)-v_k(x)|. $$ The right-hand side is a tail of a convergent series and tends to zero as \(n\to\infty\). Therefore \((v_k(x))\) is Cauchy in \(\mathbb{R}\) and has a finite limit. Define \(v(x)=\lim_{k\to\infty}v_k(x)\) on \(G\), and define \(v(x)=0\) on \(X\setminus G\). The pointwise limit on the measurable set \(G\), extended by zero on its complement, is measurable. In particular, \(v_k\to v\) almost everywhere.
On \(G\), the telescoping estimate also gives $$ |v(x)|\leq |v_1(x)|+S(x). $$ Since \(v_1,S\in L^p\), this bound implies \(v\in L^p\). For example, for \(p\geq1\), $$ |v|^p\leq 2^{p-1}\bigl(|v_1|^p+S^p\bigr) $$ almost everywhere, and the right-hand side is integrable.
To prove convergence in norm, for each \(n\) define $$ T_n=\sum_{k=n}^{\infty}|v_{k+1}-v_k|. $$ This is an \(L^p\) function: the same increasing-sum argument used for \(S\), now starting at \(n\), shows $$ \|T_n\|_p\leq\sum_{k=n}^{\infty}\|v_{k+1}-v_k\|_p. $$ On \(G\), passage to the limit in the telescoping estimate yields \(|v-v_n|\leq T_n\). Hence $$ \|v-v_n\|_p\leq\|T_n\|_p \leq\sum_{k=n}^{\infty}\|v_{k+1}-v_k\|_p\longrightarrow0. $$ The last limit follows from convergence of the series of norms. This proves the lemma.
Proof of Completeness for Finite Exponents
Proof. Let \((f_n)\) be Cauchy in \(L^p\). For each positive integer \(k\), the Cauchy condition provides an integer \(N_k\) such that $$ \|f_n-f_m\|_p<2^{-k} $$ whenever \(n,m\geq N_k\). Choose indices recursively so that \(n_1\geq N_1\), and, after choosing \(n_k\), choose $$ n_{k+1}\geq\max\{N_k,N_{k+1},n_k+1\}. $$ Then \(n_k,n_{k+1}\geq N_k\), so $$ \|f_{n_{k+1}}-f_{n_k}\|_p<2^{-k}. $$ Represent each \(f_{n_k}\) by a measurable function. The selected functions have first term in \(L^p\), and their successive differences satisfy $$ \sum_{k=1}^{\infty}\|f_{n_{k+1}}-f_{n_k}\|_p \leq\sum_{k=1}^{\infty}2^{-k}=1. $$ The Summable-Difference Criterion therefore supplies \(f\in L^p(X,\mu)\) such that \(f_{n_k}\to f\) in \(L^p\).
The original sequence is Cauchy, and a subsequence of it converges to \(f\). By the proposition “A Convergent Subsequence Determines a Cauchy Sequence,” the full sequence converges to \(f\) in the metric \(d_p(g,h)=\|g-h\|_p\). Thus every \(L^p\)-Cauchy sequence has a limit in \(L^p\), which proves completeness.
The construction does not depend on the measure space having finite measure, nor does it require pointwise convergence of the original sequence. The summably controlled subsequence is what provides an almost-everywhere limit; the metric-space argument then returns from that subsequence to the original Cauchy sequence.
Worked Examples
Worked Example: Why the Initial \(L^p\) Hypothesis Matters
On \(\mathbb{R}\) with Lebesgue measure, let \(v_k(x)=1\) for every \(x\in\mathbb{R}\) and every \(k\). Each successive difference is zero, so $$ \sum_{k=1}^{\infty}\|v_{k+1}-v_k\|_p =\sum_{k=1}^{\infty}0=0. $$ Nevertheless, for any finite \(p\geq1\), $$ \int_{\mathbb{R}}|v_1(x)|^p\,dx =\int_{\mathbb{R}}1\,dx=\infty. $$ Thus \(v_1\notin L^p(\mathbb{R})\), and the pointwise limit, which is the constant function \(1\), is not in \(L^p(\mathbb{R})\) either.
This example shows why a bound on the successive differences cannot replace the requirement \(v_1\in L^p\). The differences determine how far the functions move from one another, but say nothing about whether their common starting level has finite \(L^p\) norm.
Worked Example: A Geometric Series on a Finite-Measure Set
Let \(E\in\mathcal{F}\) satisfy \(0<\mu(E)<\infty\), and define $$ v_n=\sum_{k=1}^{n}3^{-k}\mathbf{1}_E. $$ The first term \(v_1=\frac13\mathbf{1}_E\) belongs to \(L^p\), and for each \(k\geq1\), $$ v_{k+1}-v_k=3^{-(k+1)}\mathbf{1}_E. $$ Consequently, $$ \|v_{k+1}-v_k\|_p =3^{-(k+1)}\mu(E)^{1/p}. $$ The sum of these norms is finite, since $$ \sum_{k=1}^{\infty}3^{-(k+1)}\mu(E)^{1/p} =\frac{\mu(E)^{1/p}}{6}. $$ The pointwise limit is \(v=\frac12\mathbf{1}_E\), because \(\sum_{k=1}^{\infty}3^{-k}=\frac12\). The exact norm error is $$ \|v-v_n\|_p =\mu(E)^{1/p}\sum_{k=n+1}^{\infty}3^{-k} =\frac{\mu(E)^{1/p}}{2\cdot3^n}\longrightarrow0. $$ This illustrates both the almost-everywhere limit and the norm convergence asserted by the criterion.
Worked Example: Partial Sums of Shrinking Indicators
On \((0,1)\) with Lebesgue measure, fix \(1\leq p<\infty\) and set $$ w_k=2^{-k}\mathbf{1}_{(0,1/k)},\qquad V_n=\sum_{k=1}^{n}w_k. $$ The initial term belongs to \(L^p\), and the successive difference is \(V_{k+1}-V_k=w_{k+1}\). Its norm is $$ \|V_{k+1}-V_k\|_p =\|w_{k+1}\|_p =2^{-(k+1)}(1/(k+1))^{1/p}. $$ Since \((k+1)^{-1/p}\leq1\), these norms satisfy $$ \sum_{k=1}^{\infty}\|V_{k+1}-V_k\|_p \leq\sum_{k=1}^{\infty}2^{-(k+1)}<\infty. $$ For each fixed \(x\in(0,1)\), the condition \(x<1/k\) holds for only finitely many positive integers \(k\). Thus \(V_n(x)\) eventually stops changing, and its limit is $$ V(x)=\sum_{\{k:\,x<1/k\}}2^{-k}. $$ The criterion guarantees \(V\in L^p(0,1)\) and \(\|V_n-V\|_p\to0\). More specifically, the proof gives the bound $$ \|V-V_n\|_p \leq\sum_{k=n+1}^{\infty}2^{-k}k^{-1/p} \leq\sum_{k=n+1}^{\infty}2^{-k} =2^{-n}. $$ The final expression tends to zero, verifying norm convergence quantitatively.
Why the Argument Works
The summable-difference condition performs two jobs. First, Minkowski’s inequality bounds the \(L^p\) norms of finite sums of absolute differences. Monotone Convergence then transfers that uniform bound to the infinite sum \(S\), showing that the total pointwise movement is finite almost everywhere. Second, the tail of that same sum controls the distance from a selected term to the limit. Its \(L^p\) norm tends to zero because the numerical series of successive-difference norms converges.
A common pitfall is to infer that an almost-everywhere limit belongs to \(L^p\) just because it is measurable. The proof needs the initial function in \(L^p\) as well as the \(L^p\) control of the differences: together they give \(|v|\leq|v_1|+S\) almost everywhere. Another pitfall is to stop after finding a convergent subsequence. It is the Cauchy property of the original sequence that makes the subsequence limit the limit of the entire sequence.
Check Your Understanding
Use the summable-difference criterion and the completeness proof to answer the following questions.
- Where does the proof use the hypothesis \(v_1\in L^p\), and why can the successive-difference condition alone not replace it?
- Why does the Monotone Convergence Theorem apply to the finite sums \(S_N^p\)?
- How does the tail \(T_n\) provide an upper bound for \(\|v-v_n\|_p\)?
- How are the indices \(n_k\) chosen so that the successive differences of the subsequence have summable norms?
- Why does convergence of the selected subsequence imply convergence of the original sequence?