So it is pretty obvious that the limit of measurable random variables is also measurable if they converge...but I'm not sure it's true of almost sure convergence. Suppose there is a probability space over three states $\omega_1, \omega_2, \omega_3$, but with $\mathbb{P}[\omega_1] = 1$ and $\mathbb{P}[\omega_2]=\mathbb{P}[\omega_3] = 0$. Then define $X_n$ to be the random variable where $X_n(\omega_1) = 1$, $X_n(\omega_2) = 1$, and $X_n(\omega_3) = 1$, which converges almost surely to $X$ where $X(\omega_1) = 1$, $X(\omega_2) = X(\omega_3) = 2$. Let $\mathcal{F}$ be the trivial $\sigma$-algebra. Then $X_n$ is $\mathcal{F}$-measurable but $X$ is not...so how come so many textbooks say that $X$ must also be $\mathcal{F}$-measurable, even from just almost sure convergence?

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    $\begingroup$ it's measurable wrt to the sigma field of the measurable sets. $\endgroup$ – mm-aops Oct 3 '14 at 21:48
  • $\begingroup$ It's not though because $\{\omega \in \Omega | X(\omega) \le 1\} = \{\omega_1\}$ which is not in $\{\Omega, \emptyset\}$. $\endgroup$ – Alan Faith Oct 3 '14 at 21:53
  • $\begingroup$ the trivial sigma field is not the field of measurable sets. all sets of measure $0$ are always measurable. observe there's no hope in obtaining measurability for an arbitrary sigma field since the notion of convergence you have in mind (almost sure) is directly related to the particular measure you're using, hence also to the sigma field associated to it $\endgroup$ – mm-aops Oct 3 '14 at 22:39

Random variables are not really functions on the probability space, but rather equivalence classes of such functions, where two functions are deemed equivalent if they are equal almost surely. So while the function $X$ is not $\mathcal F$-measurable, it is equal almost surely to one that is.


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