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Let $B(r)$ designate the open ball of radius $r>0$, and a fixed center in $\mathbb{R}^{n}$. Let $\lambda$ be the Lebesgue measure of dimension $n$, and $f$ be a positive locally integrable function. My question is: is it true that $$\frac{1}{\lambda(B(r))}\int_{B(r)}fd\lambda\leq \frac{1}{\lambda(B(R))}\int_{B(R)}fd\lambda,$$ if $0<r<R$? Why?

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No: define $f(x)=1+1_{B(r)}(x)$. Then $$ \frac{1}{\lambda(B(r))}\int_{B(r)}f(x)\;dx=2 $$ while $$ \frac{1}{\lambda(B(R))}\int_{B(R)}f(x)\;dx=\frac{\lambda(B(r))}{\lambda(B(R))}+1=\Big(\frac{r}{R}\Big)^n+1<2 $$

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Let $f$ be your favorite continuous, strictly positive, integrable function (e.g. Gaussian density); say $\int_{\mathbb{R}^n} f\,d\lambda = 1$. Then for $R \gg 1$ the right side is approximately $\frac{1}{R^n} \approx 0$. But for $r \ll 1$ the left side is approximately $f(0) > 0$ (where I'm taking $0$ to be the center of the ball).

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