Lebesgue integral question Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a measurable function. Show that if $f$ is continuous and fix $x_0\in \mathbb{R}$, ￼then $\lim_{n\rightarrow\infty}n\int_{x_0}^{x_0+1/n}fdm=f(x_0)$. 
Hint: Use the max-min theorem which says: For any continuous function $f$ on a compact set $[a,b]$, there are points $x_{\max},x_{\min}\in[a,b]$ such that $f(x_{\max})\geq f(x)\geq f(x_{\min})$ for all $x\in[a,b]$.
 A: The function is continuous, so we may just use the fundamental theorem of calculus for the Riemann integral, which of course agrees with the Lebesgue one.
By the FTC, 
$$
\lim_{h\to 0}\frac{1}{h}\int_{x_0}^{x_0+h}f(x)\mathrm dx=f(x_0)
$$
This will certainly hold along your particular sequence $h=1/n$.
If you want to work harder, for a given $\epsilon>0$, find $\delta>0$ with 
$$
x_0-\delta<x<x_0+\delta\implies f(x_0)-\epsilon<f(x)<f(x_0)+\epsilon
$$
then for $n>1/\delta$, and with monotonicity of integration
$$
n\int_{x_0}^{x_0+1/n}f(x_0)-\epsilon<n\int_{x_0}^{x_0+1/n}f(x)dx<n\int_{x_0}^{x_0+1/n}f(x_0)+\epsilon\\
\implies f(x_0)-\epsilon<n\int_{x_0}^{x_0+1/n}f(x)dx<f(x_0)+\epsilon
$$
A: Unnecessarily sophisticated proof that uses only continuity at $x_0$: start with the change of variable $x = x_0 + s/n$,
$$n\int_{x_0}^{x_0+1/n}f(x)\,dx = \int_0^1 f(x_0 + s/n)\,ds$$
and apply the dominated convergence theorem (the continuity of $f$ in $x_0$ implies that the functions $s\mapsto f(x_0 + s/n)$ are uniformly bounded for $n$ large enough).
Less sophisticated variant: the functions $s\mapsto f(x_0 + s/n)$ converge uniformly to the constant $f(x_0)$.
