# If $\{f_j\}$ is a sequence of measurable functions, then $\sup_j f_j(x)$ is measurable.

Similar questions are asked in math.SE but what I am especially interested is not asked (as far as I see).

If $\{f_j\}$ is a sequence of $\overline{\mathbb{R}}$-valued measurable functions on $(X,\mathcal{M})$, then $g_1(x) = \sup_j f_j(x)$ (and in fact $g_2(x) = \inf_j f_j(x)$) is measurable.

This is a proposition in Folland, Real Analysis and its proof as follows.

We have $$g_1^{-1}((a,\infty]) = \bigcup_1^{\infty}f_j^{-1}((a,\infty])$$ and $$g_2^{-1}([-\infty,a))=\bigcup_1^{\infty}f_j^{-1}([-\infty,a))$$

so $g_1$ and $g_2$ are measurable.

What I do not understand is, how can we convert the inverse of supremums and infimums to unions of the sets as done in above?

Hint: if $g_1 \geq a$ then at least one of $f_j \geq a$ – Ilya Jan 10 at 17:51
You are asking about why $sup_i f_i(x) > a \Leftrightarrow f_i(x) > a$ for some $i$. Can you prove this yourself, or are you looking for intuition? – Sanchez Jan 10 at 19:22