6
$\begingroup$

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?
$\endgroup$
2
  • 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

9
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.