# Corollary of theorem 1.14 Rudin real and complex, supremum of a sequence of measurable functions.

Theorem 1.14 states:

If $f_n:X\rightarrow [-\infty, \infty]$ is measurable, for $n = 1,2,3, ...,$ and $$g = \sup_{n \ge 1} f_n, \ h = \lim_{n \rightarrow \infty} \sup f_n$$

then g and h are measurable.

A corollary is that if $f$ and $g$ are measurable (with range in $[-\infty, \infty]$), then so are max{f,g} and min{f,g}.

I would not know how to prove this corollary given theorem 1.14, could I have a proof?

Given $f,g$ measurable, call $f_1 = f$, $f_2 = f_3 = f_4 =\dots = g$. Then clearly $$\max \{ f,g \} = \sup_n f_n$$ so you can apply the theorem.

For the minimum note that $\min \{ f,g\} = - \max \{ -f, -g\}$.

EDIT: Given two functions $f,g: X \longrightarrow [-\infty, +\infty]$ the function $m = \max\{f,g \}$ is defined pointwise $$m(x) = \max \{ f(x), g(x)\}$$ so it is a particular case of the supremum of a sequence of functions, where in the sequence there are only two functions, namely $f$ and $g$.

• But there is an infinitude of elements in max{f,g} right? so how do we know that the maximum exists? – Monolite Feb 10 '15 at 16:53
• Oh, so I think your problem is the definition of $\max \{ f,g\}$. See the edit in the answer. – Crostul Feb 10 '15 at 16:55
• I don't follow your answer - what do you mean by $f_2=f_3=\cdots=g$? – Math1000 Feb 10 '15 at 17:50
• @Math1000 They mean $f_i = g$ for all $i \ge 2, i \in \Bbb N$. – epsilon-emperor Mar 1 at 5:00

Let $g_n=\sup_{1\leqslant k\leqslant n}f_k$, then for each $n$, $g_{n+1}=\max\{g_n, f_{n+1}\}$. By induction you can show that $g_n$ is measurable for each $n$. From there you can show that $g$ is measurable as the limit of a monotone increasing sequence of measurable functions.

Similarly, let $h_n = \inf_{1\leqslant k\leqslant n}g_n$, then $h_{n+1}=\min\{h_n, g_{n+1} \}$, and use an analogous argument.