Lebesgue's Differentiation Theorem for Continuous Functions

Question

If $f:\mathbb{R}^n\to\mathbb{R}$ is continuous, does Lebesgue's differentiation theorem hold at all points? That is, does $$\lim_{r\to0}\frac{1}{|B(x,r)|}\int_{B(x,r)}f(y) \, dy=f(x)$$ $\textit{everywhere}$?

@vvnitram It actually holds. If $f$ is continuous, everything goes well (either by the FTC, or -- more "funny and suboptimal" -- by inspecting the proof and noting that if $f$ is continuous, then $f=g$...) — Clement C., May 14 '16 at 19:14

score 15 · Accepted Answer · answered May 14 '16 at 19:12

15

Yes. Fix $x\in\mathbb{R}^n$ and $\varepsilon>0$, and choose $\delta>0$ such that if $|x-y|<\delta$ then $|f(x)-f(y)|\leq\varepsilon$. If $0<r<\delta$, then $$\Big|\frac{1}{|B(x,r)|}\int_{B(x,r)}f(y)\;dy-f(x)\Big|\leq \frac{1}{B(x,r)}\int_{B(x,r)}|f(x)-f(y)|\;dy\leq\varepsilon$$ which shows that $$ \lim_{r\to 0}\frac{1}{|B(x,r)|}\int_{B(x,r)}f(y)\;dy=f(x)$$

answered May 14 '16 at 19:12

carmichael561

53,688

Nice proof - thanks! – Dan Burrows May 14 '16 at 19:28
What's the estimate done in the first part? – mavavilj Oct 17 '19 at 20:50
Which part specifically? – carmichael561 Oct 17 '19 at 23:44
When you draw the upper bound for $\bigg| \frac{1}{|B(x,r)|} \int_{B(x,r)} f(y)dy-f(x) \bigg|$. Particularly, why can you bring f(x) into the integral? – mavavilj Oct 18 '19 at 07:51
1

Because the average value of a constant $c$ over $B(x,r)$ is just $c$. In this case the constant is $f(x)$. – carmichael561 Oct 18 '19 at 23:39
@carmichael561 it is not obvious for me that $f(x)$ is constant. Can you explain in some way why we say that it is constant? When I see $f(x)$ it is stange for me too say it is constant. Does it depend to the fact that in this particular moment we look at $f(x)$ as at the limit? – Mokhmad-Salekh Khekhaev Sep 10 '21 at 13:09

score 1 · Answer 2 · answered May 14 '16 at 19:12

1

Note that near $x$ $|f(x)-f(y)|\le f(x)+\epsilon(y)$ with $\epsilon(y)\to 0$ as $y\to x$ by definition of continuity, in particular it is a bounded function. But then we have

$${1\over\operatorname{vol}(B(x,r))}\left|\int_{B(x,r)}f(y)\,dy -\int_{B(x,r)}f(x)\right| \le {1\over\operatorname{vol}(B(x,r))} \left| \int_{B(x,r)}\epsilon(y)\,dy\right|$$

$$\le \sup_{y\in B(x,r)}\epsilon(y)\stackrel{y\to x}{\longrightarrow}0$$

answered May 14 '16 at 19:12

Adam Hughes

36,777

Thanks Adam, I struggle with these sort of arguments so it's helpful to see another example of one. – Dan Burrows May 14 '16 at 19:29

Lebesgue's Differentiation Theorem for Continuous Functions

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