# If $f_k \to 0$ a.e. and $\sum_n n 2^n \mu\{|f_k| \in (2^{n-1}, 2^n]\} \leq 1$ for all $k$, then $\int f_k \to 0$.

(Stanford Real Analysis Qualifying Exam: Spring 2012) (Ideal time: 18 minutes)

1. (a) Let $\mu$ denote Lebesgue measure on $[0,1]$. Let $f_k\colon [0,1] \to \mathbb{R}$ be Lebesgue measurable functions, $k \in \mathbb{N}$. Let $m_{kn} = \mu\{x \in [0,1]\colon |f_k(x)| \in (2^{n-1}, 2^n]\}$.

Suppose that $\sum_{n=1}^\infty n2^nm_{kn} \leq 1$ for all $k \in \mathbb{N}$, and that $f_k \to 0$ a.e. Show that $\int f_k \,d\mu \to 0$.

My attempt:

Let $E_{kn} = \{x \in [0,1]\colon |f_k(x)| \in (2^{n-1}, 2^n]\}$. Then \begin{align*} \int_0^1 f_k & = \int_{\{|f_k| \leq 1\}} f_k + \int_{\{|f_k| > 1\}} f_k \\ & = \int_{\{|f_k| \leq 1\}} f_k + \sum_{n=1}^\infty \int_{E_{kn}} f_k \\ & \leq \mu\{|f_k| \leq 1\} + \sum_{n=1}^\infty 2^n m_{kn}. \end{align*} However, I'm stuck at this point.

For what it's worth, I've also thought about using the Dominated Convergence Theorem, but I don't quite see how.

• Can you prove that $\sum_{n=1}^{\infty}\int_{E_{k_n}}f_k \to 0$ by hypothesis? And $\int_{|f_k|\le 1}f_k \to 0$ by DCT? – Coiacy Mar 23 '13 at 9:07
• @Coiacy: I see that $\int_{|f_k| \leq 1} f_k \to 0$ by DCT. But I'm not sure why $\sum_{n=1}^\infty \int_{E_{kn}}f_k \to 0$. – Jesse Qual Prep Mar 23 '13 at 19:14

The splitting $\{|f| \leq 1\}$ vs $\{|f| > 1\}$ is way too crude. One would better write, $$\int f_k\,du = \int f_k\,1_{\{|f_k| \leq 2^N\}}\,d\mu + \int f_k\,1_{\{|f_k| > 2^N\}}\,d\mu$$

First integral $\int f_k\,1_{\{|f_k| \leq 2^N\}}\,d\mu$

It converges to $0$ by DCT, for every $N$.

Second integral.

We have $$\left|\int f_k1_{\{|f_k| > 2^N\}}\,d\mu\right|\leq\sum_{n>N} \int 2^n 1_{\{2^{n-1}<|f_k|\leq 2^n\}}\,d\mu = \sum_{n > N} 2^n m_{k,n}$$ and we use a nice trick: $$\sum_{n>N} 2^n m_{k,n} \leq \sum_{n>N} \frac{n}{N}2^n m_{k,n} \leq \frac{1}{N}$$

Conclusion

Taking limits as $k\to\infty$, we have $$\limsup_{k} \left|\int f_k \,d\mu\right| \leq \frac{1}{N}$$ Finally, let $N \to +\infty$.