I am going through the proof of the following.

Let $(X,\mu)$ be a measure space and $f\colon X\to\overline{\mathbb R}$ be a measurable function with finite integral. If $A_1,A_2,\dots$ are $\mu$-measurable and $\lim_{n\to\infty}\mu(A_n)=0$, then $$\lim_{n\to\infty}\int_{A_n}f\,d\mu=0.$$


Let $E_N=\{x\mid f(x)\leq N\}$ and define $f_N=\chi_{E_N}f$. Clearly $f_N\leq f$ and since $f$ has finite integral it is finite a.e., and it follows that $\lim_{N\to\infty} f_N=f$ a.e.; then, $\lim_{N\to\infty}(f-f_N)=0$ a.e.. Now $f-f_N$ is dominated by $|f|+|f_N|$ which has finite integral by additivity and hence by the DCT we get $\int (f-f_N)\,d\mu\to0$ as $N\to\infty$.

Now by additivity and the fact that $f_N\leq N$ for all $N$ we have $$\int_{A_n}f\,d\mu=\int_{A_n}(f-f_N)\,d\mu+\int_{A_n} f_N\,d\mu\leq \int_{A_n}(f-f_N)\,d\mu+N\mu(A_n)$$

Taking $N\to\infty$ and noting that $\int_{A_n}(f-f_N)\,d\mu\leq \int(f-f_N)\, d\mu$ gives $$\int_{A_n}f\,d\mu\leq 0+\infty\cdot\mu(A_n).$$ Taking $n\to\infty$ gives the result because $\infty\cdot 0=0$.

Now I don't really see how the last line proves anything. How do we know that $f$ is not negative? As I see, we have only shown that the limit is not positive. It also makes me a little uncomfortable that we write $\infty\cdot\mu(A_n)$, I mean I know that in the extended reals $0\cdot \infty=0$ but is this really rigorous?

Any ideas?

  • 3
    $\begingroup$ The use of $\infty$ there is unacceptable. $\endgroup$ – user99914 Apr 15 '16 at 18:58
  • $\begingroup$ Can't you bound the integral by product or the value of the integral of $f$ on $\mathbb{R}$ with the measure of each $A_n$, which will go to $0$ as $n\to\infty$? $\endgroup$ – John Martin Apr 15 '16 at 19:23
  • $\begingroup$ In symbols: $\int_{A_n}f \leq \mu(A_n)\int_{\mathbb{R}}f$, and the integral of $f$ is finite, say it's equal to $M$ so you have that $\mu(A_n)M\to 0$ as $n\to\infty$. It's been a while since I have thought about this stuff, and I have left some details out but something like that should work... $\endgroup$ – John Martin Apr 15 '16 at 19:25

Following the notation of your proof, set $E_N = \{ x \in X \: \colon \vert f(x) \vert \leq N\}$ and $f_N = f \cdot \chi_{E_N}$ for all $N \in \mathbb{N}$. Then $\lim_{N \to \infty} f_N = f$ almost everywhere and $f_N \leq f$ for all $N$. Since $f$ is integrable, we can apply the DCT and find \begin{equation} \lim_{N \to \infty} \int \big| f - f_N \big| \, \textrm{d} \mu = 0. \end{equation} Now, using that $\vert f_N \vert \leq N$, we find that \begin{equation}\label{eq1} \Bigg| \int_{A_n} f \, \textrm{d} \mu \Bigg|\leq \int_{A_n} \big| f - f_N \big| \, \textrm{d}\mu + N \mu(A_n) \leq \int \big| f-f_N \big| \, \textrm{d} \mu + N \mu(A_n). \end{equation} As $\mu (A_n)$ tends to $0$ as $n \to \infty$, we obtain \begin{equation} \limsup_{n \to \infty} \Bigg| \int_{A_n} f \, \textrm{d} \mu \Bigg| \leq \int \big| f - f_N \big| \, \textrm{d}\mu + N \cdot 0 = \int \big| f - f_N \big| \, \textrm{d}\mu. \end{equation} This inequality holds for every $N \in \mathbb{N}$; in particular, it remains true in the limit $N \to \infty$, and hence: \begin{equation} \limsup_{n \to \infty} \Bigg| \int_{A_n} f \, \textrm{d} \mu \Bigg| \leq 0. \end{equation} Thus, \begin{equation} 0 \geq \limsup_{n \to \infty} \Bigg| \int_{A_n} f \, \textrm{d} \mu \Bigg| \geq \liminf_{n \to \infty} \Bigg| \int_{A_n} f \, \textrm{d} \mu \Bigg| \geq 0. \end{equation} This proves the claim.


I agree with @JohnMa that the proof is not rigorous since, by definition,

$$\infty \cdot \mu(A_n) = \begin{cases} \infty, & \mu(A_n)>0, \\ 0, & \mu(A_n) \end{cases}$$

and so the estimate just shows $\int_{A_n} f \, d\mu \leq \infty$.

Without loss of generality, we may assume $f \geq 0$ (otherwise replace $f$ by $|f|$). Fix $R>0$, then

$$\int_{A_n} f \, d\mu = \int_{A_n \cap \{f \leq R\}} f \, d\mu + \int_{A_n \cap \{f>R\}} f \, d\mu$$


$$\int_{A_n} f \, d\mu \leq R \mu(A_n) + \int_{\{f>R\}} f \, d\mu.$$


$$\limsup_{n \to \infty} \int_{A_n} f \, d\mu \leq \int_{\{f>R\}} f \, d\mu$$

for any $R>0$. Letting $R \to \infty$, it follows from the dominated convergence theorem (or monotone convergence theorem) that

$$\limsup_{n \to \infty} \int_{A_n} f \, d\mu \leq 0.$$

Since $f$ is non-negative, this shows

$$\lim_{n \to \infty} \int_{A_n} f \, d\mu = 0.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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