# 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 $$x\in[a,b]$$.

• Try something more elementary. No Theorem from measure theory used to interchange limits and integrals is required. – Kavi Rama Murthy Nov 14 '18 at 6:13
• I suggest you look at the definition of a Riemann integral. Try showing this result for a very simple partition. – Jabbath Nov 14 '18 at 6:14
• Think of applying mean value theorem on the function given by the indefinite integral of $f$, which is differentiable with derivative $f$. – астон вілла олоф мэллбэрг Nov 14 '18 at 6:18

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 $$x_0-\delta then for $$n>1/\delta$$, and with monotonicity of integration $$n\int_{x_0}^{x_0+1/n}f(x_0)-\epsilon
Unnecessarily sophisticated proof that uses only continuity at $$x_0$$: start with the change of variable $$x = x_0 + s/n$$, $$n\int_{x_0}^{x_0+1/n}f(x)\,dx = \int_0^1 f(x_0 + s/n)\,ds$$ 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 $$f(x_0)$$.