It is easy to show that $$ \lim_{n \rightarrow \infty} n\int_0^\frac{1}{n} x~\mathrm{d}x. $$ I was wondering whether given $g \in C(\mathbb{R}) \cap L^1(\mathbb{R})$, it also holds that $$ \lim_{n \rightarrow \infty} n\int_{\left \lbrace x \in \mathbb{R}: 0 \leq g(x) \leq \frac{1}{n} \right \rbrace} g(x)~\mathrm{d}x = g(0) $$ We can assume that $\left\lbrace x \in \mathbb{R}: g(x) = 0 \right \rbrace$ has finite measure.
I tried to substitute $g(x) = y$ but since I do not know the structure of $g$, this does not help a lot. I know it looks a lot like Lebesgue's differentiation theorem, but I do not know any version that I can apply. Any help/reference or hint is appreciated. I would also be happy if you could tell me if there is another class of functions $g$ for which this holds.
Not true. Take $$ g(x) = \min(1, \frac1{x^2}). $$ Then the integral in question is equal to $$ 2n \int_{\sqrt n}^\infty \frac1{x^2} dx = 2n \frac1{\sqrt n} \to\infty. $$ The limit has nothing to do with Lebesgue differentiation. It is more a question on how much mass of $g$ is on sets, where $g$ is small.