# $\lim_{h\downarrow 0}\frac{1}{h}\int_{t-h}^tf(s)ds=f(t)$

Let $f:\mathbb{R}\to\mathbb{R}$ be Borel and bounded, then I was able to prove that the map $t\mapsto \int_{t-h}^tf(s)ds$ is Lipschitz continuous. Now if we assume in addition that $f$ is left-continuous then

$$\lim_{h\downarrow 0}\frac{1}{h}\int_{t-h}^tf(s)ds=f(t)$$

What I've done: let $r:=\frac{1}{h}(s-t)+1$ such that

$$\lim_{h\downarrow 0}\frac{1}{h}\int_{t-h}^tf(s)ds=\lim_{h\downarrow 0}\int_0^1f(h(r-1)+t)dr$$

Since $f$ is bounded I can use Dominated convergence to obtain

$$\int_0^1\lim_{h\downarrow 0}f(h(r-1)+t)dr=f(t)\int_0^1dr=f(t)$$

Is this correct?

However, I don't see the importance of left-continuity. Wouldn't be the same true for a right continuous $f$ and changing the integral boundaries:

$$\lim_{h\downarrow 0}\frac{1}{h}\int_{t}^{t+h}f(s)ds=f(t)$$

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It would be the same, with $h\downarrow 0$ in your last formula. – Etienne Feb 28 '14 at 10:55
@Etienne thanks for the comment. edited the last line. – math Feb 28 '14 at 16:12

I think that using Dominated convergence theorem is an overkill; my preference is to use more rudimentary tools when possible. Fix $t$. By the definition of left-continuity, for every $\epsilon>0$ there is $\delta>0$ such that $|f(s)-f(t)|<\epsilon$ whenever $s\in (t-\delta,t)$. Therefore, for $0<h<\delta$ we have $$\frac{1}{h}\int_{t-h}^t f(s)\,ds \le \frac{1}{h}\int_{t-h}^t (f(t)+\epsilon)\,ds = f(t)+\epsilon$$ and similarly $$\frac{1}{h}\int_{t-h}^t f(s)\,ds \ge \frac{1}{h}\int_{t-h}^t (f(t)-\epsilon)\,ds = f(t)-\epsilon$$ By the definition of a limit, $\lim_{h\downarrow 0 }\frac{1}{h}\int_{t-h}^t f(s)\,ds =f(t)$.