Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

Thanks in advance!

share|improve this question
Hint: if $g_1 \geq a$ then at least one of $f_j \geq a$ – Ilya Jan 10 at 17:51
3  
@Ilya I think you want strict inequalities there. – David Mitra Jan 10 at 17:53
OK but I did not gain too much unfortunately. – Mark Jan 10 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? – Sanchez Jan 10 at 19:22
No, I understand that what Ilya said. But now I understand with in the relation with the proposition. Thanks. – Mark Jan 10 at 20:01
show 1 more comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.