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Let $f_n: E \to \mathbb{R} \cup \{\infty\}$ be Leb.-integrable and suppose:

1) There is a sequence $\{a_n\}$ s.th. $a_n \ge 0$

2) $\sum_{n=1}^{\infty} a_n = L$ (i.e.: it converges to some L)

3)$\int_E |f_n(x)| \le a_n$ for all $n$

I want to show that $f_n \to 0$ almost everywhere.

My initial approach in solving this has been to note that \begin{equation} 0 \le \int_E |f_n | \le a_n \end{equation}

And then to apply a summation to the above inequality from $1$ to $n$. After doing this, I can use the Monotone Convergence Theorem so I can work with $\int_E \sum_{i=1}^{n} |f_i| \le \sum_{i=1}^{n} a_i $.

At this point, I'm thinking I can take the limit as n goes to $\infty$, however this is where I'm hitting a wall. Everything thus far seems intuitive, but I wonder if I'm missing something or incorrectly incorporating the above sums (perhaps in how I indexed?). Also, I know from previous theorems in this course that since $f$ is Leb.-integrable, $|f_n|$ is as well, and functions that are leb.-integrable are finite a.e (when working on the extended reals, as I am here). I believe this will come into play at some point in this proof, but where at I'm unsure.

Would a good strategy be to show, somehow, that $\sum_{I=1}^{n} \int_E |f_i|$ converges to $0$ a.e., and using this show that $f_n \to 0$?

Any clarification will be welcomed.

Tadpole
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3 Answers3

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Since $\sum_{n=1}^{\infty}a_n <\infty$, we have that $$a_n\to 0$$ Then by sandwich rule, you deduce that $$\int_E|f_n| \to 0$$ But, this doesn't quite give you that $f_n\to 0$ yet - do you know the Borel-Cantelli lemma?

Tyr Curtis
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  • I don't know the lemma, but after a quick search I'm guessing that the $\sum_n \int_E |f_n|$ being finite is similar to $\sum_{n=1}^{\infty} P(A_n)$ where, by the lemma, $P(A_n) = 0$. In the context of my question, I could use this lemma to show that $\int_E |f_n| = 0$, and by the theorem I mentioned earlier, $f_n = 0$ a.e. Is this correct? And if so, is it the same as saying $f$ converges to $0$ a.e.? Also, does this work in the context of only the Lebesgue integral? Are there more elementary approaches? – Tadpole May 03 '15 at 18:57
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Given $\epsilon > 0$ and $n \in \Bbb N$, let $A_n(\epsilon) := \{x\in E: |f_n(x)| \ge \epsilon\}$. Then $$m(A_n(\epsilon)) \le \frac{1}{\epsilon} \int_E |f_n(x)|\, dx \le \frac{a_n}{\epsilon} .$$

Hence, if $A(\epsilon) = \cup_{n = 1}^\infty A_n(\epsilon)$, then

$$m(A(\epsilon)) \le \sum_{n = 1}^\infty m(A_n(\epsilon)) \le \frac{L}{\epsilon}.$$

Now, the sets $\cup_{k\ge n} A_k(\epsilon)$, $n\ge 1$, are decreasing in $n$ and contained in $A(\epsilon)$, a set with finite measure. So by continuity of $m$ from above,

$$m(\limsup_{n\to \infty} A_n(\epsilon)) = \lim_{n\to \infty} m\left(\bigcup_{k \ge n} A_k(\epsilon)\right) \le \lim_{n\to \infty} \sum_{k \ge n} m(A_k(\epsilon))= 0.$$

Thus, $\limsup_{n\to \infty} A_n(\epsilon)$ has measure zero for all $\epsilon > 0$.

Let $X$ be the set of points $x \in E$ for which $\lim\limits_{n\to \infty} f_n(x) \neq 0$. Then $X = \cup_{j \ge 1} \limsup\limits_{n\to \infty} A_n(j)$, a countable union of sets of measure zero. So $X$ has measure zero, which implies $f_n \to 0$ almost everywhere.

kobe
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The monotone convergence theorem shows

$$ \int_E \sum_n |f_n| = \sum_n \int_E |f_n| \leq \sum_n a_n < \infty. $$

Now, an integrable function (with values in $[0,\infty]$) is finite almost everywhere.

Why exactly does that show your claim?

PhoemueX
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