Let $g: \mathbb{R} \rightarrow \mathbb{R^{+}}$ be any measurable function and for any $\epsilon\geq0$, let $B_{\epsilon}= \{ x\in\mathbb{R}\:\vert\: g(x)>\epsilon\}$. Now show that $$\underset{n\to\infty}{\lim}\lambda(B_{1/n})=\lambda(B_{0})$$
I'm pretty lost on this one. How can we even be sure that $g^{-1}(\{g(x)>0\}$ is a Lebesgue measurable set? Any tips/hints are much appreciated.
Let $f_n(x)$ be the function which is 1 if $x\in B_{1/n}$ and zero otherwise. Then $f_n(x)$ is nondecreasing in $n$. Further, the pointwise limit of $f_n$, call it $f$, is the indicator of $B_0$. So the monotone convergence theorem gives $$ \lim_n \lambda(B_{1/n})=\lim_n \int f_n = \int \lim_n f_n=\int f=\lambda(B_0). $$
Note that $B_0 = \cup_n B_{1 \over n}$, and the $B_{1 \over n}$ are non-decreasing. Hence $\lambda B_0 = \lim_{n \to \infty} \lambda B_{1 \over n}$.
More detail: The result follows from countable additivity on disjoint sets. Let $\Delta_1 = B_1$ and $\Delta_{n} = B_{1 \over n} \setminus B_{1 \over n-1}$. Then $B_0 = \cup_n \Delta_n$ and the $\Delta_n$ are disjoint. Hence $\lambda B_0 = \sum_n \lambda \Delta_n$. Since $\sum_{k=1}^n \Delta_k = B_{1 \over n}$, we have the desired result.