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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 $x\in[a,b]$.

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  • $\begingroup$ Try something more elementary. No Theorem from measure theory used to interchange limits and integrals is required. $\endgroup$ – Kavi Rama Murthy Nov 14 '18 at 6:13
  • $\begingroup$ I suggest you look at the definition of a Riemann integral. Try showing this result for a very simple partition. $\endgroup$ – Jabbath Nov 14 '18 at 6:14
  • $\begingroup$ Think of applying mean value theorem on the function given by the indefinite integral of $f$, which is differentiable with derivative $f$. $\endgroup$ – астон вілла олоф мэллбэрг Nov 14 '18 at 6:18
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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 $$ n\int_{x_0}^{x_0+1/n}f(x_0)-\epsilon<n\int_{x_0}^{x_0+1/n}f(x)dx<n\int_{x_0}^{x_0+1/n}f(x_0)+\epsilon\\ \implies f(x_0)-\epsilon<n\int_{x_0}^{x_0+1/n}f(x)dx<f(x_0)+\epsilon $$

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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 $f(x_0)$.

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