I am working through some revision notes, and I have come across this proof in my lecture notes that the class of $\mathcal{F}$-measurable functions is closed under limit operations:

Let $\{f_n\}$ be a sequence of $\mathcal{F}$-measurable functions on $\Omega$. Recall our definitions of $\limsup$ and $\liminf$, and apply them here pointwise to our sequence of functions $\{f_n\}$. That is, $(\limsup_{n\to\infty} f_n)(x) = \sup_{m\geqslant n}\{f_m(x)\}$.

Now let $g_n(x)=\sup_{m\geqslant n}\{f_m(x)\}$, so that $(g_n)$ is a pointwise increasing sequence, and $g_n\uparrow g$, where $g=\limsup_{n\to\infty} f_n$. Similarly define $(h_n)$ to be the pointwise decreasing sequence where $h_n(x)=\inf_{m\geqslant n}\{f_m(x)\}$, so that $h_n\downarrow h$, where $h = \liminf_{n\to\infty} f_n$.

For any $a\in\mathbb{R}$ we have $$ \{g_n \leqslant a\} = \bigcap_{m=n}^{\infty} \{f_m \leqslant a\} $$ which implies that $g_n$ is $\mathcal{F}$-measurable.

I don't understand where the conclusion in the last line comes from though, since nowhere have I seen any properties of countable intersections concerning measurable spaces. Am I missing something very obvious, or is there maybe another line that I could add in to make the conclusion a bit more clear?

Edit: Since I am not too sure how commonplace this notation is, for $f\colon\Omega\to\mathbb{R}$, $\{f < a\}=\{\omega\in\Omega \colon f(\omega) < a\}$

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    $\begingroup$ By applying De Morgan's laws, you can show straight from the definition of a $\sigma$-algebra that countable intersections of measurable sets are measurable. $\endgroup$ – fuglede Apr 8 '14 at 13:35

Here $g_{n}\left(\omega\right)=\sup\left\{ f_{m}\left(\omega\right)\mid m\geq n\right\} $ for each $\omega\in\Omega$ right?

Then $g_{n}\left(\omega\right)\leq a$ if and only if $f_{m}\left(\omega\right)\leq a$ for each $m\geq n$.

You could also say that $\omega\in\left\{ g_{n}\leq a\right\} $ if and only $\omega\in\cap_{m\geq n}\left\{ f_{m}\leq a\right\} $.

This is true for every $\omega\in \Omega$ so that means exactly that the sets $\left\{ g_{n}\leq a\right\} $ and $\cap_{m\geq n}\left\{ f_{m}\leq a\right\} $ are the same.

It shows that set $\left\{ g_{n}\leq a\right\} $ is a countable intersection of measurable sets, and this implies that the set is measurable.


Characteristic for a $\sigma$-algebra is that it is closed under complements and countable unions. Then it must also be closed under countable intersections. This because: $$\cap_{n=1}^{\infty}A_{n}=\left(\cup_{n=1}^{\infty}A_{n}^{c}\right)^{c}$$

  • $\begingroup$ That last point was the one that I was unclear about: a countable intersection of measurable sets is measurable. It seems sensible, but I've not seen it mentioned before, and would not have any idea on how to start a proof of that. Any way you could explain it a bit? $\endgroup$ – Tim Apr 9 '14 at 14:43
  • $\begingroup$ Is my addendum sufficient as explanation? $\endgroup$ – drhab Apr 9 '14 at 15:40
  • $\begingroup$ Ah yes, glaringly obvious in hindsight. Thanks very much! $\endgroup$ – Tim Apr 9 '14 at 19:44

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