Passage to the limit when the integral is infinity?

Question

Let $B(a,r)$ be a ball in $\mathbb{R}^{n}$, $n\geq2$. Suppose $f$ is measurable and $$\int_{B(a,R)}f(x)dx=\infty$$ for all $R>r>0$. Can we say that $$\int_{B(a,r)}f(x)dx=\infty?$$

Answer 1

Let $a=0$, $f(x)=0$ for $\|x\|\leq r$ and $f(x)=\frac 1 {\|x\|^{n-1} (\|x\|-r)}$ for $r <\|x\| <\infty$. Use polar coordinates in $\mathbb R^{n}$ to show that this serves as a counterexample. [When $n=2$ you can use ordinary polar coordinates. For $n >2$ see 'polar coordinates' in the index of Rudin's RCA].

Answer 2

No, take $f=\infty \chi_{B(a,r)^c}$. If $R>r$, then $m(B(a,R)\setminus B(a,r))>0$ so $\int_{B(a,R)} f=\infty$ but $\int_{B(a,r)} f=\infty m(B(a,r)\cap B(a,r)^c)=0$.

If $f$ must be in $\Bbb R$, put $f=(|x-a|-r)^n\chi_{\bar B(a,r)^c}$ and define $E_k(R)=B(a,r+\frac R{k})\setminus B(a,r+\frac R{k+1})$ for $R>r$ so that $\frac {k^n}{R^n} < f$ on $E_k(R)$. Because $B(a,R)\cap \bar B(a,r)^c=\bigcup_{k\ge 1} E_k(R)$ and the $E_k(R)$ are disjoint, we have $$\begin{align}\int_{B(a,R)} f &= \sum_{k=1}^{\infty} \int_{E_k(R)} f \ge \sum_{k=1}^{\infty} \int_{E_k(R)} \frac {k^n}{R^n}= R^{-n}\sum_{k=1}^{\infty} m(E_k(R))k^n \\ &=R^{-n}\sum_{k=1}^{\infty} \left(m\left(B\left(a,r+\frac R{k}\right)\right)-m\left( B\left(a,r+\frac R{k+1}\right)\right)\right)k^n \\ &=CR^{-n}\sum_{k=1}^{\infty} \left(\left(r+\frac R{k}\right)^n-\left( r+\frac R{k+1}\right)^n\right)k^n\end{align}$$ Where $C$ depends only on the dimension of $\Bbb R^n$. Comparing the last summand to $\frac 1k$ and applying the limit comparison test shows that $\int_{B(a,R)} f$ diverges with the harmonic series even though $\int_{B(a,r)}f =0$.