sups and infs of measurable functions are measurable

[[concept]]

Topics

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Theorem

Let $E\subset \mathbb{R}$ be measurable and $f_n:E\to [-\infty ,+\infty ]$ a sequence of measurable functions. Then the functions

• $g_1(x)={\sup }_n{f}_n(x)$
• $g_2(x)={\inf }_n{f}_n(x)$
• $g_3(x)=\lim {\sup }_{n\to \infty }{f}_n(x)={\inf }_n[{\sup }_{k\ge n}{f}_k(x)]$
• $g_4(x)=\lim {\inf }_{n\to \infty }{f}_n(x)={\sup }_n[{\inf }_{k\ge n}{f}_k(x)]$ Are all measurable functions

Proof

$g_1(x)={\sup }_n{f}_n(x)$ Note that

x \in g_{1}^{-1}(\,(\alpha, \infty]\,) &\iff \sup_{n} f_{n}(x) > \alpha \\ & \iff \exists\, n \text{ s.t. }f_{n}(x) > \alpha \\ &\iff x \in f_{n}^{-1} (\,(\alpha, \infty]\,) \\ &\iff x \in \bigcup_{n} f_{n}^{-1}(\, (\alpha, \infty]\,) \end{align}$$ And each set in this union is [[Concept Wiki/lebesgue measurable\|measurable]] (since each $f_{n}$ is [[Concept Wiki/measurable function\|measurable]]), and thus the preimage of $(\alpha, \infty]$ under $g_{1}$ is also measurable. Thus $g_{1}$ is measurable.

$g_2(x)={\inf }_n{f}_n(x)$ In a similar manner (and using our equivalent intervals), we can check that $x\in g_2^{-1}([\alpha ,\infty ])⟺x\in {\bigcap }_n{f}_n^{-1}([\alpha ,\infty ])$ And since each $f_n^{-1}([\alpha ,\infty ])$ is measurable, we have that the countable intersection is integrable, and therefore that $g_2$ is also measurable.

Finally, $g_3$ and $g_4$ follow immediately from the two proofs above showing measurable closure under both infimums and supremums (since they are the sup and inf of a sequence of measurable functions respectively).

References

References

See Also

Mentions

Mentions

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