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.



About mathtuition88

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