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Let $f_m(x)$ be a sequence of measurable functions. Prove that: $$F^*(x)=\inf_{n\geq 1}\sup_{m\geq n}f_m(x)$$ is also measurable.

I have the definition of a measurable function is given as:

Function $f:X\to \mathbb R$ is said to be measurable if $\forall \alpha \in \mathbb R \{x\in X:f(x)<\alpha\}\in \mathcal{X},$ where $\mathcal{X}$ is the given sigma algebra.

I can prove that $\sup_{n \geq 1}f_n(x)$ and $\inf _{n \geq 1}f_n(x)$ are measurable functions since $$H_n(x)=\sup_{n \geq 1}f_n(x)=\bigcup_{n=1}^{\infty}\{x\in X:f_n(x)>\alpha\}$$

and that :

$$G_n(x)=\inf_{n \geq 1}f_n(x)=\bigcap_{n=1}^{\infty}\{x\in X:f_n(x)>\alpha\}$$

Can I then say that $F^*(x)$ is measurable because $$F^*(x)=\inf_{n\geq1}{H_n(x)}$$ is the infinum of measurable functions, and ive already proved that the infinum of measurable functions is measurable?

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  • $\begingroup$ How could $H_n(x)=\bigcup_{n=1}^{\infty}\{x\in X:f_n(x)>\alpha\}$? Something is definitely incorrect there. If you correct it then your proof should work out just fine. $\endgroup$ – BigbearZzz Aug 2 '16 at 13:36
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Let $g_n(x)=\sup_{m\geq n}f_m(x)$. Since $$\sup_{m\geq n} u_m\leq a\iff \forall m\geq n, u_m\leq a,$$ $$\{x\mid g_n(x)\leq a\}=\bigcap_{m\geq n}\{x\mid f_m(x)\leq a\},$$ and thus $g$ is measurable. Since $(g_n)$ is decreasing, $$F^*(x)=\lim_{n\to \infty }g_n(x),$$ and thus $F^*$ is measurable.

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