Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mu$ be a probability measure on $X$, so that $\int_X \mu(dx) = 1$.

Say under which conditions on the function $f: X \rightarrow \mathbb{R}_{> 0} \ $ (that is measurable and integrable) we have that

$$ \lim_{\mu(A) \rightarrow 0 } \int_A f(x) \mu(dx) = 0 $$

share|cite|improve this question
up vote 4 down vote accepted

In fact these conditions (measurable and integrable) are already sufficient. Indeed, let $A$ measurable. We have for a fixed $n$, denoting $E_n:=\left\{x,f(x)\leq n\right\}$: \begin{align} \int_A f(x)d\mu(x)&=\int_{A\cap E_n} f(x)d\mu(x)+\int_{A\cap E_n^c} f(x)d\mu(x)\\ &\leq n\mu(A)+\int_{E_n^c} f(x)d\mu(x)\\ &\leq n\mu(A)+\sum_{k=n}^{+\infty}(k+1)\mu(k\leq f < k+1),
\end{align} and since the series $\sum_{k=1}^{+\infty}k\mu(k\leq f<k+1)$ is convergent, so is the series $\sum_{k=1}^{+\infty}(k+1)\mu(k\leq f<k+1)$, hence we can, given $\varepsilon>0$, find a $n$ such that $\sum_{k=n}^{+\infty}(k+1)\mu(k\leq f<k+1)\leq \frac{\varepsilon}2$. Then for each $A$ measurable such that $\mu(A)\leq \frac{\varepsilon}{2n}$, we have $\int_A f(x)d\mu(x)\leq \varepsilon$.

share|cite|improve this answer
Can you please explain your second inequality? And also the passage "since" [...] is convergent, so is the series [...]? – Adam Mar 10 '12 at 0:33
I write $E_n^c$ has the union $\bigcup_{k\geq n}\{x,k\leq f(x)<k+1\}$, and on this set $f(x)<k+1$. We have $\sum_{k=1}^{+\infty}k\mu(k\leq f<k+1)\leq \int f d\mu<\infty$ and the series $\sum_{k=0}^{+\infty}\mu(k\leq f<k+1)$ is convergent since it's the measure of the set $\{f\geq 0\}$, which is finite since the measure of $X$ is finite. – Davide Giraudo Mar 10 '12 at 9:55

Another way to see this is to note that if $A_n$ is a sequence of measurable sets with $\mu(A_n) \to 0$, then $f 1_{A_n} \to 0$ in measure. Since $|f 1_{A_n}| \le |f|$ and $f$ is integrable, an appropriate version of the dominated convergence theorem shows that $\int 1_{A_n} f\,d\mu \to 0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.