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I'm trying to do a proof $f(x)=\lim_{r\to0^+}\frac{1}{2r}\int_{x-r}^{x+r} f(y)dy$ for all x when $f:\mathbb{R}\to\mathbb{R}$ and $f$ is continous.

I would like to use continuity and not the fundamental theorem of calculus. I assume it should be possible, but I don't really even know where to start.

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  • $\begingroup$ If you do not want to use the fundamental theorem of calculus, you will have to start with the definition of a (Riemann) integral. $\endgroup$ – Fabian Mar 22 '17 at 10:52
  • $\begingroup$ @Fabian Not necessarily, the property that $g\leqslant h$ pointwise implies $\int_a^bg\leqslant\int_a^bh$ suffices. $\endgroup$ – Did Mar 22 '17 at 11:57
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You might consider the squeeze lemma for integrals: if $g(x) \le f(x) \le h(x)$ for all $x$, then $\int_a^b g(t) dt \le \int_a^b f(t) dt \le \int_a^b h(t) dt$. This is particularly useful when the functions $g$ and $h$ vary with some parameter, and as they do so, their integrals get closer and closer.

In this case, let's fix $x$. Because $f$ is continuous, on the interval $[x-r, x+r]$, $f$ has a maximum value $M_r$ and a minimum value $m_r$. Hence

$$ \int_{x-r}^{x+r} m_r~ dx \le \int_{x-r}^{x+r} f(x)~ dx \le \int_{x-r}^{x+r} M_r ~dx $$ which simplifies down to $$ 2rm_r \le \int_{x-r}^{x+r} f(x)~ dx \le 2rM_r $$ or $$ m_r \le \frac{1}{2r}\int_{x-r}^{x+r} f(x) ~dx \le M_r $$ Now if you can show that $\lim_{r \to 0^{+}} m_r = \lim_{r \to 0^{+}} M_r$, you'll be done.

Can you do that? Hint: $f$ is continuous at $x$, and we really haven't used that very strongly yet.

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We can write $$ \int_{x-r}^{x+r}f(y)\,dy= \int_{x-r}^{x}f(y)\,dy+ \int_{x}^{x+r}f(y)\,dy $$ Thus, with the simple change of variables $r\mapsto -r$, it is easy to prove that $$ \lim_{r\to0^+}\frac{1}{r}\int_{x-r}^{x}f(y)\,dy= \lim_{r\to0^-}\frac{1}{r}\int_{x}^{x+r}f(y)\,dy $$ (provided one of them exists).

Now, the fact that $$ \lim_{x\to0}\frac{1}{r}\int_{x}^{x+r}f(y)\,dy=f(x) $$ is the fundamental theorem of calculus, so in order to avoid it you have essentially to write down a proof of the theorem, which is what the other answers do.

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By the mean value theorem (for integrals) there is $t_r \in (x-r,x+r)$ such that

$\frac{1}{2r}\int_{x-r}^{x+r} f(y)dy=f(t_r)$

Since $t_r \to x$ for $r \to 0^+$, the continuity of $f$ gives

$\frac{1}{2r}\int_{x-r}^{x+r} f(y)dy \to f(x) $for $r \to 0^+$.

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Let $\varepsilon>0$. Since $f$ is continuous, there is a $r>0$ with $|f(x)-f(y)|<\varepsilon$ for all $y\in[x-r,x+r]$ resp. $f(y)-\varepsilon<f(x)<f(y)+\varepsilon$ for all y.. With monotonicity of the integral follows $$\int_{x-r}^{x+r}(f(y)-\varepsilon)dy<\int_{x-r}^{x+r}f(x)dy<\int_{x-r}^{x+r}(f(y)+\varepsilon)dy$$ which leads to $$\int_{x-r}^{x+r}f(y)dy-2r\varepsilon<2rf(x)<\int_{x-r}^{x+r}f(y)dy+2r\varepsilon.$$ Now divide by $2r$ and you obtain $$\frac{1}{2r}\int_{x-r}^{x+r}f(y)dy-\varepsilon<f(x)<\frac{1}{2r}\int_{x-r}^{x+r}f(y)dy+\varepsilon$$ resp $$\left|f(x)-\frac{1}{2r}\int_{x-r}^{x+r}f(y)dy\right|<\varepsilon.$$ Note that you can choose $r$ w.l.o.g. arbitrarily small.

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