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

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

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.
 
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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.