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Assume that we are given two random sequences, defined on the same probability spaces and taking values from $\ell_{2}$. Next, assume that $X_{n}\overset{a.s.}{\to} a$ and $Y_{n}\overset{a.s.}{\to} b$.

Next, let's takes some continuous function $f: \ell_{2} \times \ell_{2} \to \mathbb{R}$. Can we now apply continuous mapping theorem to prove that $$ f(X_{n}, Y_{n}) \overset{a.s.}{\to} f(a,b)? $$

Attempt: if $f$ is just the inner product, then the result it correct, see, for example Convergence in inner product space.

Also, here Functional continuous mapping theorem is the example where one must be careful with continuous mapping theorem.

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  • $\begingroup$ Your notion of convergence means that for almost all $\omega$ in the probability space you want $X_n(\omega)\to a$ where $a\in \ell_2$? $\endgroup$ – s.harp May 6 at 8:47
  • $\begingroup$ yes, the standard meaning on a.s. convergence in Probability theory $\endgroup$ – LrM May 6 at 14:25
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The set of $\omega$ for which either $X_n(\omega)$ or $Y_n(\omega)$ doesn't converge is a null set, as it is a union of two null sets.

So for almost all $\omega$ you have $(X_n(\omega), Y_n(\omega))\to (a,b)$. By continuity of $f$ you then get that for such $\omega$ $f(X_n(\omega),Y_n(\omega))\to f(a,b)$ - hence $f(X_n, Y_n)$ converges almost surely to $f(a,b)$.

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  • $\begingroup$ this is the thing: I did not say that $X_{n}$ and $Y_{n}$ converge together as random element element of product space $\ell_{2} \times \ell_{2}$. Though, both $X_{n}$ and $Y_{n}$ are defined on the same probability space. $\endgroup$ – LrM May 6 at 15:29
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    $\begingroup$ @LrM: It should be easy to show that if $X_n \to a$ a.s. and $Y_n \to b$ a.s, then $(X_n, Y_n) \to (a,b)$ in the product topology a.s. $\endgroup$ – Nate Eldredge May 6 at 15:55
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    $\begingroup$ @LrM the set of $\omega$ for which $X_n(\omega)$ doesn't converge to $a$ is a null set. The set of $\omega$ for which $Y_n(\omega)$ doesn't converge to $b$ is a null set. The union of those two sets is then a null set - and the union is precisely the set of points $\omega$ for which $(X_n(\omega) , Y_{n}(\omega))$ doesn't converge to $(a,b)$. $\endgroup$ – s.harp May 6 at 16:04

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