changing measurable functions on a measure zero set preserves measurability

[[concept]]

Topics

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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 $N=\{x\in E:f(x)\ne g(x)\}$. By assumption, this set has outer measure zero. Thus $m(N)=0$. So for all $\alpha \in \mathbb{R}$, we have $N_{\alpha }=\{x\in N:g(x)>\alpha \}\subset N$ Also has measure zero (since $m^{\ast }(N_{\alpha })\le m^{\ast }(N)=0$) and is therefore measurable.

Now, the preimages of $f$ and $g$ are the same outside of $N$, and thus we get $g^{-1}((\alpha ,\infty ])=(f^{-1}((\alpha ,\infty ])\cap N^c)\cup N_{\alpha }$ 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. \tag*{$\blacksquare$}

References

References

See Also

Mentions

Mentions

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