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Let $f: [a, b] \to \mathbb R$ continuous function. Show that for all $x \in (a, b)$ applies that

$$\lim_{\delta \to 0} {1 \over 2\delta} \int_{x-\delta}^{x+\delta} f(t) dt = f(x)$$

Does the same result apply to all integrable functions $f: [a, b] \to \mathbb R$?

I cant figure out what property of continuous function should I use for this and how, also I would appreciate a hint for the other part as well.

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For the first, select $\epsilon > 0$. Then pick $\delta$ small enough so that $|t-x|<\delta\implies |f(t)-f(x)|<\epsilon$. Then consider

$${1\over 2\delta}\left|\int_{x-\delta}^{x+\delta}f(t)-f(x)\,dt\right|\le {1\over 2\delta}\int_{x-\delta}^{x+\delta}|f(t)-f(x)|\,dt \le\epsilon$$

showing the result for continuous functions.

Now let

$$g(x) = \begin{cases} 1 & |x|\le 1, x\ne 0 \\ 0 & x=0\end{cases}$$

Then, with $x=0$, we have

$$\displaystyle{1\over 2\delta}\int_{-\delta}^{\delta}g(t)\,dt = 1\ne 0=g(0).$$

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    So I break the $ {1\over 2\delta}\int_{x-\delta}^{x+\delta}|f(t)-f(x)|\,dt $ in to two parts (one with $f(t)$ and the other with $f(x)$ and get to the point where $\lim_{\delta \to 0} {1 \over 2\delta} \int_{x-\delta}^{x+\delta} f(t) dt - f(x) \le \epsilon$? and that's the wanted result? – 2017-02-23
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    @repulsive23 note that $\displaystyle {1\over 2\delta}\int_{-\delta}^\delta f(x)\,dt=f(x){2\delta\over 2\delta}=f(x)$ because $f(x)$ is a **constant** when integrated with respect to $t$, so you can say $${1\over 2\delta}\int_{-\delta}^{\delta}\big(f(t)\,dt\big)-f(x) = {1\over 2\delta}\int_{-\delta}^{\delta}\big(f(t)-f(x)\,dt\big).$$ – 2017-02-23
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    Yes, that is exactly what I meant. – 2017-02-23
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    @repulsive23 then yes, you have exactly divined what I was saying. Cheers! – 2017-02-23