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So I have stumbled upon the following theorem:

Let $\left\{f_n\right\}$ be a sequence of measurable functions. For $x \in X$, put $$ g(x) = \sup \left\{ f_n (x) \mid n \in \mathbb{N} \right\} \\ h(x) = \limsup_{n \to \infty} f_n (x) $$ Then $g$ and $h$ are measurable

The proof is as follows:

$$ \left\{ x \mid g(x) > a \right\} = \bigcup_{n=1}^{\infty} \left\{ x \mid f_n (x) > a \right\} $$

At least to me, this seems to not be true, so I was wondering if someone could help me understand where I go awry.

It seems to me that you could have a sequence of functions $f_n (x) = x + \frac{x}{n} : n \neq 3 , f_3 (x) = 1000$. From the proof $$ g(x) = \sup \left\{ x + \frac{x}{n} \mid n \in \mathbb{N} \right\} = \left\{ \begin{array}{rr} 2x & : x \ge 0 \\ x & : x < 0 \end{array} \right. $$ Now it seems to me that each $f_n (x)$ is measurable in $\mathbb{R}$ however the statement in the proof isn't true here. Put $a = 3$ then $\left\{ x \mid g(x) > 3 \right\} = \left( \frac{3}{2}, \infty \right)$ however $\bigcup_{n=1}^{\infty} \left\{ x \mid f_n (x) > 3 \right\} = (- \infty , \infty )$ because $f_3 (x) > 3 \; \forall \; x \in \mathbb{R}$, so wouldn't this be a counter example to the proof given?

Thanks in advance!

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Haven't you completely ignored your own definition of $f_3$ here? By definition, $g(x)\geq f_3(x)=1000$ for all $x$. – Nick Peterson Jul 20 '13 at 0:40
@NicholasR.Peterson oop, I put in the wrong value, I meant a different $a$ and meant $f_3 = -1000$ – DanZimm Jul 20 '13 at 0:45
@NicholasR.Peterson you enabled me to see where I was thinking wrong, so I wanted to thank you! – DanZimm Jul 20 '13 at 0:47
up vote 6 down vote accepted

You have a sequence $\langle f_n\rangle$ and you set $$g(x)=\sup_{n\geqslant 1}\{f_n(x)\}$$

The claim is that

$$\left\{ x \mid g(x) > a \right\} = \bigcup_{n=1}^{\infty} \left\{ x \mid f_n (x) > a \right\}$$

First, suppose $x$ is such that $g(x)>a$. This means $$\sup_{n\geqslant 1}\{f_n(x)\}>a$$

This means that there exists $m$ such that $f_m(x)>a^{\dagger}$, so $x\in \{x\mid f_m(x)>a\}$ and thus $x$ is in that union. Conversely, suppose $x$ is in the union. Then for some $m$ we have $f_m(x)>a$, thus $\sup\limits_{n\geqslant 1 }\{f_n(x)\}=g(x)>a$.

$\dagger$: Note that if for each $n$ we had $f_n(x)\leq a$ then the $\sup$ would be $\leq a$. Thus, by contraposition, we must have $f_m(x)>a$ for at least one $m$.

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Ah ok thank you, I will accept once I can (4 minutes from this comments post) – DanZimm Jul 20 '13 at 0:47

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