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Let $(X,\mathcal{M},\mu)$ be a measure space and $f:X\rightarrow\mathbb{C}$ be a measurable function such that $\int_X |f|\; d\mu <+\infty$. I am trying to prove the following result:

\begin{equation} \forall \epsilon >0,\exists \delta >0, \forall E\in\mathcal{M}, \mu(E)<\delta\Rightarrow \int_E |f|\; d\mu <\epsilon. \end{equation}

I considered proving the contraposition (which semmed easier), that is if $\exists \epsilon >0$ such that $\forall\delta>0,\exists E\in\mathcal{M}$ satisfying $\mu(E)<\delta$ and $\int_E |f|\; d\mu \geq \epsilon$, then $\int_X |f|\; d\mu = +\infty$, but I cannot get to the result. It is easy to prove that in this case, $|f|$ is not bounded but it is basically the only result I have achieved. Is there a way to work things out?

TerranDrop
  • 840

3 Answers3

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If we let $f_n$ be a monotone increasing sequence of simple functions converging pointwise to $|f|$, then we can take $n$ such that $$\int_X(|f|-f_n)d\mu<\frac{\epsilon}{2}.$$ Now, $f_n$ is bounded, say by $N>0$, so if $\mu(E)<\frac{\epsilon}{2N}$ then $\int_E f_nd\mu\leq\frac{\epsilon}{2}$.

Finally, for such $E$, $$\int_E|f|d\mu = \int_E(|f|-f_n)d\mu + \int_E f_nd\mu\leq \int_X(|f|-f_n)d\mu + \int_E f_nd\mu\leq\epsilon.$$

Jonathan Y.
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My favorite way to deal with such problems is the following identity : for any $\sigma$-finite measured space $(X, \mathcal{M}, \mu)$ and any measurable, complex-valued function $f$ on this space,

$$\int_X |f| \ d \mu = \int_0^{+ \infty} \mu (|f| \ge t) \ dt. \ (1)$$

Let us denote the function $t \mapsto \mu (|f| \ge t)$ by $F$. Since $f$ is integrable on $(X, \mu)$, so is $F$ on $\mathbb{R_+}$. Then, for any $\delta \ge 0$, for any measurable subset $E \subset X$ such that $\mu (E) \le \delta$,

$$\int_E |f| \ d \mu = \int_0^{+ \infty} \mu ( \{|f| \ge t \} \cap E) \ dt \le \int_0^{+ \infty} \min ( \delta, F(t)) \ dt.$$

From there I think you can work things out. One of the advantages of this method is that it is quite general, and makes it really easy to get quantitative bounds (i.e. given $f$ and $\varepsilon >0$, to find an explicit $\delta$ which works).

Also, if you haven't seen the identity $(1)$ before, here is a hint to prove it: use Fubini's theorem on the function $1_{|f|(x) \ge t} (x, t)$ define on $X \times \mathbb{R}_+$.

D. Thomine
  • 10,870
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Continuing your idea: for all $n\in\mathbb N$, there exists $E_n$ such that $\mu(E_n)<\frac{1}{n^2}$ and $\int_{E_n}|f|\geq\varepsilon$. Then, from the first Borel-Cantelli lemma, you have that $\mu\left(\limsup_{n\to\infty}E_n\right)=0$, so $$\int_{\limsup_{n\to\infty}E_n}|f|=0.$$ Now, if $A_n=\bigcup_{m\geq n}E_m$, you have that $(A_n)$ is decreasing and $\limsup_{n\to\infty}E_n=\bigcap_{n\in\mathbb N}A_n$. Since $f$ is integrable, $\nu(A)=\int_A|f|$ defines a measure on $X$, so $$\lim_{n\to\infty}\int_{A_n}|f|=\int_{\bigcap_{n\in\mathbb N}A_n}|f|=\int_{\limsup_{n\to\infty}E_n}|f|=0.$$ But $E_n\subseteq A_n$, so $$\int_{A_n}|f|\geq\int_{E_n}|f|\geq\varepsilon,$$ which is a contradiction.

detnvvp
  • 8,237