changing measurable functions on a measure zero set preserves measurability

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Theorem

If two functions $f, g : E \to [- \infty, \infty]$ satisfy $f = g$ almost everywhere on $E$ and $f$ is measurable, then $g$ is measurable.

ie, changing a measurable function on a set of measure zero does not affect its measurability.

Proof

Let $N = {x \in E : f (x) \neq g (x)}$ . By assumption, this set has outer measure zero. Thus $m (N) = 0$ . So for all $α \in R$ , we have

N_{α} = {x \in N : g (x) > α} \subset N

Also has measure zero (since $m^{*} (N_{α}) \leq m^{*} (N) = 0$ ) and is therefore measurable.

Now, the preimages of $f$ and $g$ are the same outside of $N$ , and thus we get

g^{- 1} ((α, \infty]) = (f^{- 1} ((α, \infty]) \cap N^{c}) \cup N_{α}

But $N$ is measurable, so $N^{c}$ is also measurable. Thus all sets of RHS are measurable, and so RHS is also measurable. Thus $g$ is measurable, as desired.

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File	Last Modified
Functional Analysis Lecture 10	2025-07-14
Functional Analysis Lecture 9	2025-07-14

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Last Modified

Functional Analysis Lecture 10

2025-07-14

Functional Analysis Lecture 9

2025-07-14

changing measurable functions on a measure zero set preserves measurability

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