Here is the problem that I am dealing with specifically:
I was able to prove part (a) and (b), where in part (b), I showed that the Hausdorff Outer measure is a metric outer measure and thus the Catheodory's Set of the Hausdorff Outer measure contains all closed set and thus all Borel sets. In particular, this tells us that Hausdorff Measure is a Borel measure on $\mathbf{R}^n$ when restricted to the Borel $\sigma$-algebra by the Catheodory's Theorem.
However, part (c) seems to have assumed that Hausdorff Outer measure is a measure when restricted to the set of Lebesgue measurable sets on $\mathbf{R}^n$. Why is this true? I have no idea how to extend the measure from the Borel $\sigma$-algebra to the Lebesgue $\sigma$-algebra. I also have the potential solution saying the following about my question:
I am, however, not entirely convinced of this solution. In particular, I am not following how knowing $\lambda^*(A) = 0 \implies H^*_d(A) = 0$ gives us the Hausdorff measurability of all Lebesgue sets? Assuming this solution is correct, I would like to see a bit more details on the logic behind the sentence. Any advices would be appreciated.
A Lebesgue measurable set has the form $B\triangle N$, where $B$ is a Borel set and $N$ is a Lebesgue-null set (and $\triangle$ is symmetric difference). Then $N$ is also $H_d^*$-null and thus $H_d$-measurable, so $B\triangle N$ is $H_d$-measurable as a symmetric difference of two measurable sets.