Let $R$ be a rectangle in $\mathbb{R}^n$. Let $f:R\to\mathbb{R}$ be a bounded, integrable function. Suppose that $f$ is continuous at some interior point $a$ of $R$. Show that $$\lim_{r\to 0^+} \frac{1}{\text{vol}(B_r(a))} \int_{B_r(a)} f\, dV=f(a).$$ Note:

  1. $\int_{B_r(a)} f\, dV$ is defined as $\int_{R_1}\bar{f}\, dV$ for some rectangle $R_1\supset B_r(a)$, where $\bar{f}:\mathbb{R}^n\to\mathbb{R}$ is defined by $\bar{f}=f$ on $B_r(a)$ and $\bar{f}=0$ otherwise.
  2. $\text{vol}(B_r(a))$ is defined as $\int_{B_r(a)}1\,dV$.

My attempt:

Fix $\epsilon>0$. Choose $\delta>0$ such that $|f(x)-f(a)|<\epsilon_1$ (later defined) for all $x\in B_\delta(a)\subset R$. Fix $r\in(0,\delta)$. Let $I:=\int_{B_r(a)}f\, dV$. For any partition $P$ of $R_1$, we have $$ L(1,P)\le \text{vol}(B_r(a)) \le U(1_, P)$$ $$ L(\bar{f},P)\le I\le U(\bar{f},P) $$

Suppose that $f>0$ on $B_\delta(a)$. Then $$ L(\bar{f},P)/U(1, P) \le I/\text{vol}(B_r(a))\le U(\bar{f},P)/L(1,P) $$ We have $ U(\bar{f},P)/L(1,P) = \sum_{Q\in P}\sup(\bar{f}(Q))\text{vol}(Q)/\sum_{Q\in P}\inf(1(Q))\text{vol}(Q) $.

I do not know how to proceed after expanding the definition.


1 Answer 1


You don't have to expand into lower and upper sums. Looking at the integral and the triangle inequality suffices. We get $$\left|\frac{1}{\operatorname{vol}(B_r(a))}\int_{B_r(a)}fdV-f(a)\right|=\left|\frac{1}{\operatorname{vol}(B_r(a))}\int_{B_r(a)}fdV-\frac{f(a)\operatorname{vol}(B_r(a))}{\operatorname{vol}(B_r(a))}\right|=$$$$\left|\frac{1}{\operatorname{vol}(B_r(a))}\int_{B_r(a)}f-f(a)dV\right|\leq\frac{1}{\operatorname{vol}(B_r(a))}\int_{B_r(a)}|f-f(a)|dV<$$$$\frac{1}{\operatorname{vol}(B_r(a))}\int_{B_r(a)}\varepsilon dV=\frac{\operatorname{vol}(B_r(a))}{\operatorname{vol}(B_r(a))}\varepsilon=\varepsilon.$$


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