sups and infs of measurable functions are measurable

Theorem

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

$g_{1} (x) = sup_{n} f_{n} (x)$
$g_{2} (x) = inf_{n} f_{n} (x)$
$g_{3} (x) = lim sup_{n \to \infty} f_{n} (x) = inf_{n} [sup_{k \geq n} f_{k} (x)]$
$g_{4} (x) = lim inf_{n \to \infty} f_{n} (x) = sup_{n} [inf_{k \geq n} f_{k} (x)]$
Are all measurable functions

Proof

g_{1} (x) = sup_{n} f_{n} (x)

Note that

\begin{aligned} x \in g_{1}^{- 1} ((α, \infty]) & ⟺ sup_{n} f_{n} (x) > α \\ ⟺ \exists n s.t. f_{n} (x) > α \\ ⟺ x \in f_{n}^{- 1} ((α, \infty]) \\ ⟺ x \in ⋃_{n} f_{n}^{- 1} ((α, \infty]) \end{aligned}

And each set in this union is measurable (since each $f_{n}$ is measurable), and thus the preimage of $(α, \infty]$ 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} ([α, \infty]) ⟺ x \in ⋂_{n} f_{n}^{- 1} ([α, \infty])

And since each $f_{n}^{- 1} ([α, \infty])$ 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).

File	Last Modified
Functional Analysis Lecture 10	2025-07-14
Functional Analysis Lecture 9	2025-07-14
positive and negative part of a function	2025-07-14

sups and infs of measurable functions are measurable

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