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Let $(\Omega, \mathcal{E}, \mathbb{P})$ be a probability space, and $X_n$ $(n=1,2,\dots)$ a sequence of random variables, and $g: \mathbb{R}^2 \to \mathbb{R}$ a measurable function, and $h: \mathbb{R} \to \mathbb{R}$ a real-valued function.

Are the following two statements different?

(a) $g(y, X_n) \to h(y)$ almost-surely $\mathbb{P}$, pointwise in $y$.

(b) $g(y, X_n) \to h(y)$ pointwise in $y$, almost surely $\mathbb{P}$.

I think that (a) means:

$$ \forall y. \mathbb{P}[\omega \in \Omega \mid g(y, X_n(\omega)) \to h(y)] = 1. $$

whereas (b) means: $$ \mathbb{P}[ \omega \in \Omega \mid \forall y. g(y, X_n(\omega)) \to h(y)] = 1, $$

(assuming that the event in (b) is measurable for simplicity).

Although I'm still not sure what would be the difference between these two expressions, or if this interpretation makes sense. Is one implied by the other?

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  • $\begingroup$ One place this shows up is in the Glivenko-Cantelli theorem on Wikipedia: "For every (fixed) $x, F_n(x)$ is a sequence of random variables which converge to $F(x)$ almost surely by the strong law of large numbers, that is, $F_{n}$ converges to $F$ pointwise." $\endgroup$ – jII Aug 26 '18 at 20:22
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In a) there is a null set for each $y$. b) insists that there is a null set independent of $y$. The two are not equivalent. For example, let $X$ have some continuous distribution, say normal, and $X_n=X$ for all $n$. Let $g(x,y)=1$ if $x=y$, and $g(x,y)=0$ if $x\neq y$. Let $h \equiv 0$. The you can check that a) holds, but the probability in b) is $0$, not $1$!

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