Can every set be contained in a measurable set differing by a null set?

[alexfloo]

I'm trying to prove, per ex. 5 of section 2.2 of S. Berberian's Fundamentals of Real Analysis, that where $\lambda^*$ is the Lebesgue outer measure, and A_n is any sequence of (not necessarily measurable) sets of reals increasing to A, then $\lambda^*(A_n)$ increases to $\lambda^*(A)$.

As a hint, it mentions that every set is contained in a measurable set which differs from it by a null set. I considered the closure. I know that the boundary is not necessarily null (for instance, the rationals) but perhaps this cannot be the case for a nonmeasurable set.

In either case, assuming the hint, the proof is pretty trivial. I just don't really know where to start on proving the hint.

 

[morphism]

Given any subset B of R, measurable or not, one can find a measurable set M containing B with $\lambda^*(B) = \lambda(M)$. (If the outermeasure of B is infinite, take M=R; otherwise find a descending sequence U_n of open sets containing B with $\lambda(U_n) < \lambda(B) + 1/n$, and then let M be their intersection.)

The existence of such an M should help you prove your hint.