Continuity of a function defined by an integral, when the variable is in the region of integration Hi everyone: Suppose that $f$ is locally integrable in $\mathbb{R}^{n}$, $(n\geq2)$;  $B(a,r)$ is the ball of center $a$ and radius $r>0$ , and $\lambda$ is the $n$-dimensional Lebesgue measure. It seems clear that the function  $$r\mapsto\int_{B(a,r)}f(t)d\lambda(t)$$ is continuous on $(0,+\infty)$. But how would you write a rigorous proof for it? Thanks for your reply.
 A: Assume without loss of generality that $a=0$ and let $B(0,r)=B_r$. We have to prove the following: given $\epsilon>0$ there exists $\delta>0$ such that
$$
0<r<R<\delta\implies\Bigl|\int_{B_R\setminus B_r}f(t)\,d\lambda(t)\Bigr|\le\epsilon.
$$
Since $\lambda(B_R\setminus B_r)\to0$ as $r\to R$, the result is a consequence of the following general fact of Lebesgue measure (or more generally, of measure theory.)

Let $A\subset\mathbb{R}^n$ bemeasurable and $f\colon A\to\mathbb{R}$
  integrable. Given $\epsilon>0$ there exists $\delta>0$ such that $$
B\subset A \text{ measurable
},\lambda(B)<\delta\implies\Bigl|\int_Bf(t)\,d\lambda(t)\Bigr|\le\epsilon.
$$

This is clear if $f$ is bounded. For the general case we may assume $f\ge0$. Since $f$ is integrable there is a simple function $s\colon A\to\mathbb{R}$ such that $0\le s(t)\le f(t)$ and$\int_A(f(t)-s(t))d\lambda(t)<\epsilon/2$. Since $s$ is simple it is bounded, say by $C>0$. Then for any $B\subset A$ measurable with $\lambda(B)\le\epsilon/(2\,C)$
$$
\int_Bf=\int_B(f-s)+\int_Bs\le\frac\epsilon2+C\,\lambda(B)\le\epsilon.
$$
