Jtdylktuy

Let $f:mathbb{R}rightarrowmathbb{R}$ be a measurable function. Show that if $f$ is continuous and fix $x_0in mathbb{R}$, then $lim_{nrightarrowinfty}nint_{x_0}^{x_0+1/n}fdm=f(x_0)$.

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 $xin[a,b]$.

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,
$$
lim_{hto 0}frac{1}{h}int_{x_0}^{x_0+h}f(x)mathrm dx=f(x_0)
$$
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
$$
x_0-delta<x<x_0+deltaimplies f(x_0)-epsilon<f(x)<f(x_0)+epsilon
$$
then for $n>1/delta$, and with monotonicity of integration
$$
nint_{x_0}^{x_0+1/n}f(x_0)-epsilon<nint_{x_0}^{x_0+1/n}f(x)dx<nint_{x_0}^{x_0+1/n}f(x_0)+epsilon\
implies f(x_0)-epsilon<nint_{x_0}^{x_0+1/n}f(x)dx<f(x_0)+epsilon
$$

Unnecessarily sophisticated proof that uses only continuity at $x_0$: start with the change of variable $x = x_0 + s/n$,
$$nint_{x_0}^{x_0+1/n}f(x),dx = int_0^1 f(x_0 + s/n),ds$$
and apply the dominated convergence theorem (the continuity of $f$ in $x_0$ implies that the functions $smapsto f(x_0 + s/n)$ are uniformly bounded for $n$ large enough).

Less sophisticated variant: the functions $smapsto f(x_0 + s/n)$ converge uniformly to the constant $f(x_0)$.