I'm having trouble with the following proof:

enter image description here

$\color{red}{\text{This is not the end of the proof.}}$

I'm not understanding the definition of $A_k$. For instance $A_5\nsubseteq A_6$ because $A_6$ takes $\omega$ if $|X_{n\geq6}(\omega)-X(\omega)|< \epsilon$. Then I don't understand why these sets form an increasing sequence since the $(k+1)^{th}$ set doesn't contain the elements of the $k^{th}$ set.

Any help in pointing out why my reasoning is wrong will be very much appreciated.

  • 1
    On the contrary, $A_5 \subseteq A_6$ , because if some $\omega$ belongs to $A_5$, (the inequality holds for $X_{n\ge 5}$) then it also belongs to $A_6$ – leonbloy Nov 23 '14 at 3:00
up vote 2 down vote accepted

Here is an intuitive answer. Hope it helps.

You know that $X_{n}$ converges almost surely to $X$. That is, the distance between $X_{n}$ and $X$ will be, and will remain, very small for large $n$. In other words, given a small distance, say $\epsilon = 0.00012$, then there will be at least the same, if not more, amount of elements $\omega$ in your sample space $\Omega$, such that $|X_{1000}(\omega)-X(\omega)|<\epsilon$ than $|X_{10}(\omega)-X(\omega)|<\epsilon$, say. The larger you make $n$, the closer you are to $X$, and, therefore, you will be able to find more $\omega$s such that $|X_{n}(\omega)-X(\omega)|<\epsilon$ holds true for your choice of $\epsilon$.

Since $X_n$ converges to $X$ almost everywhere, there exists a set $M$ of measure 1 with every $\omega\in M$ satisfying $X_(\omega)\rightarrow X(\omega)$. This translates to the usual epsilon definition of a limit for each omega. So clearly there exists a $K$ such that $A_K$ is non empty.

Fix $\epsilon>0$. Clearly $A_k$ is increasing since for all $n\geq k$ implies for all $n\geq k+1$ (but not the other way around). Remember, $\epsilon$ is fixed so for each $\omega$ there is a $K(\omega)$ such that $\omega\in A_k$ for all $k\geq K(\omega)$.

  • A bit difficult to read for me... Thanks. – Vladimir Vargas Nov 23 '14 at 3:39

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