Outer Measure Zero implies Measurable

One quick way is to use Caratheodory’s Criterion:

Let $\lambda^*$ denote the Lebesgue outer measure on $\mathbb{R}^n$ , and let $E\subseteq\mathbb{R}^n$ . Then $E$ is Lebesgue measurable if and only if $\lambda^*(A)=\lambda^*(A\cap E)+\lambda^*(A\cap E^c)$ for every $A\subseteq\mathbb{R}^n$ .

Suppose $E$ is a set with outer measure zero, and $A$ be any subset of $\mathbb{R}^n$ .

Then $\lambda^*(A\cap E)+\lambda^*(A\cap E^c)\leq\lambda^*(E)+\lambda^*(A)=\lambda^*(A)$ by the monotonicity of outer measure.

The other direction $\lambda^*(A)\leq\lambda^*(A\cap E)+\lambda^*(A\cap E^c)$ follows by countable subadditivity of outer measure.

Author: mathtuition88

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