Consider the space $L^2(\Omega, \mathcal{F}, P)$ of square integrable random variables (more precisely, the set of all equivalent classes which appear by identifying a.s. equal square integrable random variables). This space is known to be a Hilbert space with the scalar product

$$ \langle X,Y \rangle = E(X Y). $$

Now let T $\subset R$ and $X_t \in L^2(\Omega, \mathcal{F}, P)$ for t $\in T$. Consider the set $$ L^2(\Omega, \sigma(X_t)_{t \in T}, P) := \{ X \in L^2(\Omega, \mathcal{F}, P): X \quad \text{is} \quad \sigma(X_t)_{t \in T} - \text{measurable}\}. $$

I want to show that $L^2(\Omega, \sigma(X_t)_{t \in T}, P)$ is a closed linear subspce of $L^2(\Omega, \mathcal{F}, P)$.

The part with the linear subspace is trivial since linear combinations of $\sigma(X_t)_{t \in T}$-measurable and square integrable random variables clearly maintain the same properties.

Now I want to show that the subspace is closed, i.e. if $$ \|X_n -X\| \rightarrow 0 \iff \mathbb{E}{|X-X_n|}^2 \rightarrow0 $$ for $X_n \in L^2(\Omega, \sigma(X_t)_{t \in T}, P)$ and $X \in L^2(\Omega, \mathcal{F}, P)$, then $X \in L^2(\Omega, \sigma(X_t)_{t \in T}, P).$

I think of using the fact that the for a sequence $X_n$ of $\sigma(X_t)_{t \in T}$-measurable random variables, if the pointwise limit exists, then

$$ \lim_{n \rightarrow \infty} X_n(\omega) $$

is $\sigma(X_t)_{t \in T}$-measurable. However I know this result only with the classical limit notion and not for a limit w.r.t. an arbitrary norm. Does this result hold for the $L^2$ limit?

To summarize my questions:

  1. How can one show that $L^2(\Omega, \sigma(X_t)_{t \in T}, P)$ is closed?
  2. Does that mean that $L^2(\Omega, \sigma(X_t)_{t \in T}, P)$ is also a Hilbert space? Why?
  3. If the answer to the second question is yes, then is it true that a closed linear subspace of a Hilbert space is a Hilbert space? Why?
  • 1
    $\begingroup$ The answer to a similar question here suggests to take a subsequence from $X_n$ which converges a.s. to $X$ and conclude that $X$ is measurable. But in that case we only know that the pointwise limit exists only almost surely. Isn't this a problem for measurability? $\endgroup$
    – Holden
    Sep 15, 2017 at 17:02
  • 1
    $\begingroup$ I'm sure you know this by now, but for other, future readers, yes, it is a problem. For instance, see Proposition 2.11 in Folland, which says that a completeness condition is necessary. Moreover, even if our probability space $(\Omega,\mathcal{F},P)$ is complete (an uncommon assumption in probability), that still does not imply completeness of the relevant space $(\Omega,\bigvee_t \sigma(X_t),P)$. $\endgroup$ Jun 27 at 4:38

2 Answers 2


This is true in general: if $\mathcal{G}$ is a sub-$\sigma$-field of $\mathcal{F}$, then $L^2(\Omega, \mathcal{G}, P)$ is a closed subspace of $L^2(\Omega, \mathcal{F}, P)$.

Since you are worried about null sets, we should define the subspace more carefully: a random variable $X$ (or more precisely, its equivalence class $[X]$) is defined to be in $L^2(\Omega, \mathcal{G}, P)$ iff it is square-integrable and there exists a representative of the class $[X]$ which is $\mathcal{G}$-measurable. In other words, iff there exists $X'$ which is $\mathcal{G}$-measurable and $X=X'$ a.s.

The subsequence result you mentioned in a comment (which is a standard Borel-Cantelli argument) is basically what's needed. Suppose $X_n \in L^2(\Omega, \mathcal{G}, P)$ and $X_n \to X$ in $L^2$. We may, if desired, replace each $X_n$ with an equivalent random variable $X_n'$ which is $\mathcal{G}$-measurable. By the result you quoted, there exists a subsequence $X_{n_k}$ such that $X_{n_k} \to X$ a.s. Then $X_{n_k}' \to X$ a.s. as well. Set $X' = \limsup_{k \to \infty} X_{n_k}'$. Then $X'$ is $\mathcal{G}$-measurable, since it is the pointwise (everywhere!) limsup of a sequence of $\mathcal{G}$-measurable random variables, and we have $X' = X$ a.s. So (the equivalence class of) $X$ is in $L^2(\Omega, \mathcal{G}, P)$. (Liminf would have worked just as well.)

For (2), yes, this does imply that $L^2(\Omega, \mathcal{G}, P)$ is itself a Hilbert space. Indeed, as to (3), any closed linear subspace $K$ of a Hilbert space $H$ is itself a Hilbert space. The proof is practically trivial: after all, a (real) Hilbert space is a (real) vector space equipped with an inner product $\langle \cdot, \cdot \rangle_H : H \times H \to \mathbb{R}$ which is bilinear, symmetric, and positive definite, and such that $H$ is a complete metric space under the induced norm. Now $K$ is by assumption a linear subspace, so it is a vector space (to be pedantic, the "addition" and "scalar multiplication" operations on $K$ are the restrictions of the operations from $H$, and you can verify that all the necessary axioms remain satisfied). Moreover, if we let $\langle \cdot, \cdot \rangle_K$ be the restriction of $\langle \cdot, \cdot \rangle_H$ to $K \times K$, then it is trivial to check that $\langle \cdot, \cdot \rangle_K$ is again bilinear, symmetric, and positive definite.

This leaves completeness, and note that we have not used the closedness of $K$ yet. Let $\{x_n\}$ be a Cauchy sequence in $K$. Since the $K$-norm is the restriction of the $H$-norm, this means $\{x_n\}$ is also Cauchy in $H$. By completeness of $H$, it converges to some $x \in H$, i.e. $\|x_n - x\|_H \to 0$. But since $K$ was closed, we have $x \in K$. Again, the $K$-norm is the restriction of the $H$-norm, so $\|x_n - x\|_K = \|x_n -x\|_H \to 0$. Thus $x_n \to x$ in $K$-norm, and we have shown that $K$ is complete.

The same argument shows that a closed linear subspace of a Banach space is a Banach space.

The converse is also true, and so you can prove (1) another way: you know that $L^2$ of any measure space is a Hilbert space, so apply this to the probability space $(\Omega, \mathcal{G}, P|_{\mathcal{G}})$. Any $\mathcal{G}$-measurable random variable is a fortiori an $\mathcal{F}$-measurable random variable, so there is a natural well-defined isometric inclusion of $L^2(\Omega, \mathcal{G}, P|_{\mathcal{G}})$ into $L^2(\Omega, \mathcal{F}, P)$, whose image is the subspace in question. Thus the subspace is complete, being the isometric image of a complete space, and complete subsets are closed in any metric space (easy exercise).

  • $\begingroup$ Very illuminating, thank you. $\endgroup$
    – Holden
    Sep 16, 2017 at 15:53
  • $\begingroup$ +1 That's a trick I haven't seen before! I'll have to remember it, since these sets of measure zero do seem to be kind of important in probability. Very cool. $\endgroup$ Jun 27 at 4:41

Very nice answer above! An alternative method I came up with is neither more general nor more elegant, but I hope the diversity of ideas justifies the bits of cloud memory.

Let $(X,M,\mu)$ be a measure space an $N$ a sub-$\sigma$-algebra of $M$. We will show that the collection of $L^p(X,M,\mu)$-equivalence classes for which there's some $N$-measurable representative is closed. Once we have $N$-measurable representatives $(f_n)$ converging in $L^p(X,M,\mu)$ to an $M$-measurable representative $f$, we note that the convergence, also, occurs in $\mu$-measure (by Chebyshev's inequality for $p<\infty$, but even for $p=\infty$, by uniform convergence on a set with null complement).

So, $\lim_k f_{n_k}=f$ point-wise $\mu$-almost everywhere. This, by itself, does not imply $f$ is $N$-measurable, not even if $(X,M,\mu)$ is complete! However, we note as a lemma that $$ \|h\|_{L^p(\,X,\,N,\,\mu\rvert_N)} = \|h\|_{L^p(X,M,\mu)} $$ for any $N$-measurable $h$ (for $p=\infty$, this is automatic from the definition and, for $p<\infty$, try $h$ a characteristic function). Therefore, we obtain that $\lim_{n,m} \|f_n-f_m\|_{L^p(\,X,\,N,\,\mu\rvert_N)}=0$. Hence, like before, by Chebyshev if $p<\infty$ and otherwise by being uniformly Cauchy on a set with $\mu\rvert_N$-null complement, we obtain that $(f_n)$ is Cauchy in $\mu\rvert_N$ measure, as hence is $(f_{n_k})$, too. Therefore, $\lim_\ell f_{n_{k_\ell}} = h$ $\mu\rvert_N$-almost everywhere to some $N$-measurable $h$.

At this point, $\lim_\ell f_{n_{k_\ell}} = h$ $\mu$-almost everywhere (since the set with $\mu\rvert_N$-null complement, also, has $\mu$-null complement). Additionally, $\lim_\ell f_{n_{k_\ell}} = f$ $\mu$-almost everywhere. Therefore, $f=h$ on the intersection of two "$\mu$-almost everywheres," and so $h$ is an $N$-measurable representative of the limiting equivalence class.


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