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?

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.