Completeness of Lp Spaces

Why Completeness Matters

A Cauchy sequence in a normed space has terms that become arbitrarily close to one another. Completeness says that this internal closeness is enough: the sequence converges to an element of the same space. For $L^p$, this is not immediate from the definition. The candidate limit is not given in advance, and an $L^p$-Cauchy sequence need not converge pointwise at every outcome.

The key is to select a subsequence whose successive differences are small enough to have a summable sequence of $L^p$ norms. On a probability space, those differences are also summable in $L^1$. This gives almost-sure convergence of the subsequence; a second estimate then verifies convergence in the $L^p$ norm.

Definitions and the Completeness Claim

Definition: Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space and let $1\leq p<\infty$. The space $L^p$ consists of equivalence classes, modulo equality almost surely, of measurable random variables $X$ for which $\mathbb E[|X|^p]<\infty$. Its norm is $$\|X\|_p=\big(\mathbb E[|X|^p]\big)^{1/p}.$$ A sequence $(X_n)$ is Cauchy in $L^p$ if, for every $\varepsilon>0$, there is $N\in\mathbb N$ such that $\|X_n-X_m\|_p<\varepsilon$ whenever $n,m\geq N$.

Representatives of the equivalence classes may be chosen when discussing pointwise convergence. Any limit obtained almost surely defines an equivalence class independently of the choice on a null set. We use Minkowski’s inequality and Lyapunov’s inequality from the earlier material. In particular, on a probability space, $\|Y\|_1\leq\|Y\|_p$ for $Y\in L^p$.

Theorem: For every $1\leq p<\infty$, $L^p$ is complete. Equivalently, every Cauchy sequence in the $L^p$ norm converges in that norm to an element of $L^p$.

Summable Increments Produce a Limit

We first establish the mechanism used in the completeness proof. Suppose the increments of a sequence have summable $L^p$ norms. The conclusion is stronger than just convergence in norm: the partial sums converge almost surely as well.

Proposition: Let $1\leq p<\infty$, let $X_1\in L^p$, and let $(G_k)_{k\in\mathbb N}\subset L^p$ satisfy $\sum_{k=1}^{\infty}\|G_k\|_p<\infty$. Define $S_n=X_1+\sum_{k=1}^{n-1}G_k$, with $S_1=X_1$. Then $S_n$ converges almost surely and in $L^p$ to some $S\in L^p$.

Proof. Choose measurable representatives of the $G_k$. By Lyapunov’s inequality and the monotone convergence theorem, $$\mathbb E\!\left[\sum_{k=1}^{\infty}|G_k|\right] =\sum_{k=1}^{\infty}\mathbb E[|G_k|] \leq\sum_{k=1}^{\infty}\|G_k\|_p<\infty.$$ A nonnegative random variable with finite expectation is finite almost surely. Therefore $\sum_k|G_k(\omega)|<\infty$ for almost every $\omega$, so the series defining $S_n$ converges absolutely there. Let $S$ be its limit on this full-probability set, and define it arbitrarily on the complement. It is measurable as an almost-sure limit of measurable random variables.

For any $m\in\mathbb N$, the partial tails $\sum_{k=m}^{j}G_k$ converge almost surely, as $j\to\infty$, to $S-S_m$. Fatou’s lemma and Minkowski’s inequality give $$\mathbb E[|S-S_m|^p] \leq\liminf_{j\to\infty}\mathbb E\!\left[\left|\sum_{k=m}^{j}G_k\right|^p\right] \leq\left(\sum_{k=m}^{\infty}\|G_k\|_p\right)^p.$$ The right side is finite and tends to zero as $m\to\infty$. In particular, $S-S_1\in L^p$, so $S\in L^p$ by Minkowski’s inequality. The same bound shows $\|S-S_m\|_p\to0$. Thus the convergence is both almost sure and in $L^p$. $\square$

Proof of Completeness

Starting from a Cauchy sequence, we choose a subsequence with summable successive differences. The proposition then supplies a limit and controls the subsequence in norm. The original Cauchy property transfers that convergence to the full sequence.

Proof. Let $(X_n)$ be Cauchy in $L^p$. For each $k\in\mathbb N$, the Cauchy property gives an index $N_k$ such that $$\|X_n-X_m\|_p<2^{-k}\quad\text{whenever }n,m\geq N_k.$$ Choose indices $n_k$ strictly increasing with $n_k\geq N_k$, and put $G_k=X_{n_{k+1}}-X_{n_k}$. Since both indices are at least $N_k$, $\|G_k\|_p<2^{-k}$. Consequently $\sum_k\|G_k\|_p\leq\sum_k2^{-k}<\infty$.

Apply the proposition with initial term $X_{n_1}$ and increments $G_k$. Its partial sums are exactly $X_{n_j}$ at the corresponding indices: $X_{n_1}+\sum_{k=1}^{j-1}G_k=X_{n_j}$. Thus the subsequence converges in $L^p$ to some $X\in L^p$.

To show that the full sequence converges to $X$, let $\varepsilon>0$. Choose $N$ such that $\|X_n-X_m\|_p<\varepsilon/2$ for $n,m\geq N$. Since $n_j\to\infty$ and $X_{n_j}\to X$ in $L^p$, there is a $j$ with $n_j\geq N$ and $\|X_{n_j}-X\|_p<\varepsilon/2$. For every $n\geq N$, the triangle inequality now gives $$\|X_n-X\|_p\leq\|X_n-X_{n_j}\|_p+\|X_{n_j}-X\|_p<\varepsilon.$$ Hence $X_n\to X$ in $L^p$, proving completeness. $\square$

Worked Examples

Worked Example: A Random Series with Uniformly Small Tails

On $[0,1)$ with Lebesgue probability measure, define $r_k(x)=(-1)^{\lfloor 2^k x\rfloor}$ and $X_n=\sum_{k=1}^n3^{-k}r_k$. Each $r_k$ is measurable and satisfies $|r_k(x)|=1$. For $m>n$, the triangle inequality gives, for every $x$,

|X_m(x)-X_n(x)|\leq\sum_{k=n+1}^{m}3^{-k}\leq\sum_{k=n+1}^{\infty}3^{-k}=\frac{3^{-n}}{2}.

Thus $\|X_m-X_n\|_p\leq 3^{-n}/2$ for every finite $p\geq1$, so $(X_n)$ is $L^p$-Cauchy. The same tail estimate shows that the series converges at every $x$ to $X(x)=\sum_{k=1}^{\infty}3^{-k}r_k(x)$, and $\|X-X_n\|_p\leq3^{-n}/2$. The limit is bounded by $\sum_{k=1}^{\infty}3^{-k}=1/2$, hence belongs to every finite $L^p$.

Worked Example: Approximating a Singular Function

On $(0,1]$ with Lebesgue probability measure, fix $a$ with $0<a<1/p$, and set $X_n(x)=x^{-a}\mathbf 1_{[1/n,1]}(x)$. For $m>n$, the difference is $x^{-a}$ on $[1/m,1/n)$ and zero elsewhere, up to endpoints. Therefore

\|X_m-X_n\|_p^p=\int_{1/m}^{1/n}x^{-ap}\,dx\leq\int_0^{1/n}x^{-ap}\,dx=\frac{n^{-(1-ap)}}{1-ap}.

Because $1-ap>0$, the upper bound tends to zero as $n\to\infty$, proving that $(X_n)$ is Cauchy in $L^p$. Its pointwise limit for every $x>0$ is $X(x)=x^{-a}$. The error is explicitly

\|X-X_n\|_p^p=\int_0^{1/n}x^{-ap}\,dx=\frac{n^{-(1-ap)}}{1-ap}\longrightarrow0.

In particular, $X\in L^p$, since $\int_0^1x^{-ap}\,dx=1/(1-ap)<\infty$. The restriction $a<1/p$ is essential for this calculation: at $ap\geq1$, the integral at zero is infinite.

Worked Example: A Discrete $L^p$ Limit

Let $\Omega=\mathbb N$, with $\mathbb P(\{j\})=2^{-j}$ for $j\in\mathbb N$. These probabilities sum to one. Define $X_n(j)=j$ when $j\leq n$ and $X_n(j)=0$ otherwise. For $m>n$,

\|X_m-X_n\|_p^p=\sum_{j=n+1}^{m}2^{-j}j^p\leq\sum_{j=n+1}^{\infty}2^{-j}j^p.

The series $\sum_{j=1}^{\infty}2^{-j}j^p$ converges: the ratio of its $(j+1)$-st term to its $j$-th term is $\frac12(1+1/j)^p$, which tends to $1/2<1$. Its tails therefore tend to zero, so $(X_n)$ is Cauchy in $L^p$. At each $j$, $X_n(j)\to X(j)=j$. The same tail calculation gives $\|X-X_n\|_p^p=\sum_{j=n+1}^{\infty}2^{-j}j^p\to0$, and the convergent series also verifies $X\in L^p$.

What the Proof Does—and Does Not—Say

The subsequence is chosen for a specific purpose: summable $L^p$ increments ensure that the sum of their absolute values is finite almost surely. This establishes an almost-sure limit for that subsequence. It does not assert that the original Cauchy sequence converges pointwise at every outcome, or even almost surely. Completeness concerns convergence in the $L^p$ norm; the proof uses almost-sure convergence of a selected subsequence as a tool for constructing the norm limit.

The probability-space assumption also has a concrete role. Lyapunov’s inequality gives $\|Y\|_1\leq\|Y\|_p$, which turns summability of the $L^p$ norms into integrability of $\sum_k|G_k|$. The theorem stated here covers $1\leq p<\infty$; it makes no claim about the $L^\infty$ case.

Key takeaway: To prove completeness, make the successive differences of a subsequence summable in norm. Their absolute sum is then finite almost surely, and Fatou’s lemma together with Minkowski’s inequality identifies the resulting limit as an $L^p$ limit.

Check Your Understanding

Use the definitions and estimates in this tutorial to answer the following questions.

Why does summability of $\|G_k\|_p$ imply that $\sum_k|G_k|$ is finite almost surely on a probability space?
In the completeness proof, why is it useful to choose a subsequence with $\|X_{n_{k+1}}-X_{n_k}\|_p<2^{-k}$?
How does Fatou’s lemma give a bound on $\|S-S_m\|_p$ in the summable-increments proposition?
Why does convergence of the extracted subsequence, together with the Cauchy property, imply convergence of the full sequence?
For the singular-function example, what condition on $a$ and $p$ makes the integral at zero finite?

What You'll Learn

Why Completeness Matters

Definitions and the Completeness Claim

Summable Increments Produce a Limit

Proof of Completeness

Worked Examples

Worked Example: A Random Series with Uniformly Small Tails

Worked Example: Approximating a Singular Function

Worked Example: A Discrete \(L^p\) Limit

What the Proof Does—and Does Not—Say

Check Your Understanding

Completeness of Lp Spaces

What You'll Learn

Why Completeness Matters

Definitions and the Completeness Claim

Summable Increments Produce a Limit

Proof of Completeness

Worked Examples

Worked Example: A Random Series with Uniformly Small Tails

Worked Example: Approximating a Singular Function

Worked Example: A Discrete \(L^p\) Limit

What the Proof Does—and Does Not—Say

Check Your Understanding

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