# 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?

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Hint: if $g_1 \geq a$ then at least one of $f_j \geq a$ –  Ilya Jan 10 '13 at 17:51
@Ilya I think you want strict inequalities there. –  David Mitra Jan 10 '13 at 17:53
OK but I did not gain too much unfortunately. –  oeda Jan 10 '13 at 18:38
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? –  user27126 Jan 10 '13 at 19:22
No, I understand that what Ilya said. But now I understand with in the relation with the proposition. Thanks. –  oeda Jan 10 '13 at 20:01

The equality of sets $$g_1^{-1}((a,\infty]) = \bigcup_1^{\infty}f_j^{-1}((a,\infty])\tag1$$ means $$x \in g_1^{-1}((a,\infty]) \iff x\in \bigcup_1^{\infty}f_j^{-1}((a,\infty])\tag2$$ which is the same as $$g_1(x)>a \iff \exists j \ f_j(x)>a\tag3$$ The proof of (3) is an exercise in using the definition of supremum.
Similarly, the equality of sets $$g_2^{-1}([-\infty,a))=\bigcup_1^{\infty}f_j^{-1}([-\infty,a))\tag4$$ is converted to $$g_2(x)<a \iff \exists j \ f_j(x)<a\tag5$$