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Let $X$ and ${\left( {{X_n}} \right)_{n \in \mathbb{N}}}$ be random variables on a measurable space $\left( {\Omega ,\mathcal{F}} \right)$. Show that:

1) $\left\{ {\omega \in \Omega :\mathop {\lim }\limits_{n \to \infty } {X_n}\left( \omega \right) = X\left( \omega \right)} \right\} \in \mathcal{F}$

2) For $\Omega = \mathbb{R}$ and $X$ differentiable show that ${X'}$ is also a random variable.

My solution:

1) We know that if $X$ and $Y$ are random variables and if $g:{\mathbb{R}^2} \to \mathbb{R}$ is a Borel function, then $g\left( {X,Y} \right)$ is also a random variable. Taking $g\left( {x,y} \right) = x - y$, we know that $g$ is continuous and hence a Borel function. So $\left\{ {\omega \in \Omega :X\left( \omega \right) = Y\left( \omega \right)} \right\} = {\left( {g\left( {X,Y} \right)} \right)^{ - 1}}\left( {\left\{ 0 \right\}} \right) \in \mathcal{F}$.

If we managed to show that $Y = \mathop {\lim }\limits_{n \to \infty } {X_n}$ is a random variable, that would complete the proof. We know it's equivalent to show that $\left\{ {\omega \in \Omega :Y\left( \omega \right) \leqslant y} \right\} \in \mathcal{F},\forall y \in \mathbb{R}$ to conclude that $Y$ is a random variable. We have $$\left( {\mathop {\lim \inf }\limits_{n \to \infty } {X_n}} \right) \leqslant x \Leftrightarrow \mathop {\sup }\limits_n \left\{ {\mathop {\inf }\limits_{m \geqslant n} {X_n}} \right\} \leqslant x \Leftrightarrow \bigcap\limits_{n = 1}^\infty {\left( {\mathop {\inf }\limits_{m \geqslant n} {X_n} \leqslant x} \right)} \Leftrightarrow \bigcap\limits_{n = 1}^\infty {\bigcup\limits_{m = n}^\infty {\left( {{X_n} \leqslant x} \right)} } $$ so we conclude that ${\mathop {\lim \inf }\limits_{n \to \infty } {X_n}}$ is a random variable. By noting that $$\mathop {\lim \sup }\limits_{n \to \infty } {X_n} = \mathop {\lim }\limits_{n \to \infty } \mathop {\sup }\limits_{m \geqslant n} {X_n} = - \mathop {\lim }\limits_{n \to \infty } \mathop {\inf }\limits_{m \geqslant n} \left( { - {X_n}} \right) = - \mathop {\lim \inf }\limits_{n \to \infty } \left( { - {X_n}} \right)$$ we similarly conclude that $\mathop {\lim \sup }\limits_{n \to \infty } {X_n}$ is also a random variable.

Now, using the same argument as before, we conclude that the set $A$ consisting of all $\omega \in \Omega $ such that the limit $\mathop {\lim }\limits_{n \to \infty } {X_n}\left( \omega \right)$ exists is an event because $A = \left\{ {\mathop {\omega \in \Omega :\lim \inf }\limits_{n \to \infty } {X_n}\left( \omega \right) = \mathop {\lim \sup }\limits_{n \to \infty } {X_n}\left( \omega \right)} \right\} \in \mathcal{F}$. So, $Y$ is a random variable on $A$, since $\left\{ {\omega \in A:Y\left( \omega \right) \leqslant y} \right\} = \left\{ {\omega \in A:\mathop {\lim \inf }\limits_{n \to \infty } {X_n}\left( \omega \right) \leqslant y} \right\} \in \mathcal{F}$ for all $y \in \mathbb{R}$.

My question: What next?

2) Let ${\left( {{h_n}} \right)_{n \in \mathbb{N}}}$ be a sequence of real numbers such that $\mathop {\lim }\limits_{n \to \infty } {h_n} = 0$. Define ${X_n}\left( \omega \right): = \frac{{X\left( {\omega + {h_n}} \right) - X\left( \omega \right)}}{{{h_n}}}$. Since $X$ is differentiable, $\mathop {\lim }\limits_{n \to \infty } {X_n}\left( \omega \right)$ exists for all $\omega \in \Omega $. We conclude that $X' = \mathop {\lim }\limits_{n \to \infty } {X_n}$ is a random variable on $\Omega $ if ${X_n}$ is a random variable for all $n \in \mathbb{N}$. Since ${g_n}:\mathbb{R} \to \mathbb{R}$ defined by ${g_n}\left( x \right) = x + {h_n}$ is a continuous and hence a Borel function, it follows that $X \circ {g_n}\left( \omega \right) = X\left( {\omega + {h_n}} \right)$ is also a random variable. From here, it's easy to see that $X_n$ is also a random variable for each $n \in \mathbb{N}$.

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Simply set $$Y(\omega) = 0$$ for all $\omega \in A^c$, then $Y$ defines a random variable on $\Omega$ and the claim follows from the fact that

$$\left\{\lim_{n \to \infty} X_n = X \right\} = A \cap \{Y=X\} \in \mathcal{F}.$$


An alternative argumentation goes as follows: Note that $\lim_{n \to \infty} X_n(\omega) = X(\omega)$ if, and only if,

$$\forall k \in \mathbb{N} \, \exists N \in \mathbb{N} \, \forall n \geq N: |X_n(\omega)-X(\omega)| \leq \frac{1}{k}.$$

This is equivalent to

$$\omega \in \bigcap_{k \in \mathbb{N}} \bigcup_{N \in \mathbb{N}} \bigcap_{n \geq N} \left\{ |X_n-X| \leq \frac{1}{k} \right\}.$$

Consequently,

$$\left\{ \lim_{n \to \infty} X_n = X \right\} = \bigcap_{k \in \mathbb{N}} \bigcup_{N \in \mathbb{N}} \bigcap_{n \geq N} \left\{ |X_n-X| \leq \frac{1}{k} \right\} \in \mathcal{F}.$$

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  • $\begingroup$ I like this alternative route $\endgroup$ – Alen Feb 15 '15 at 9:47

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