Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

I found this on a qualifier exam, and I think it will help me understand $L^p$ spaces better.

Let $f_n$ be a sequence of measurable function on a finite measure space. Suppose that $$\sup_n \int_X |f_n(x)|^2 d\mu < \infty$$ and that $\lim_{n\to \infty}f_n(x) =: f(x)$ exists $\mu$-almost everywhere. Which of the following are true (proving or providing a counterexample):

(1) $\int_X |f(x)|^2 d\mu < \infty$

(2) $ \int_X |f(x)| d\mu < \infty$

(3) $\lim_{n\to\infty} \int_X |f_n(x) - f(x)|^2 d\mu = 0$

(4) $\lim_{n\to\infty} \int_X |f_n(x) - f(x)| d\mu = 0$

share|cite|improve this question
up vote 5 down vote accepted
  1. It's true by an application of Fatou's lemma to $\{|f_n|^2\}$, a sequence of non-negative measurable functions.

  2. It's true by Cauchy-Bunyakovsky-Schwarz inequality (and the fact that the measure space is finite).

  3. Take $f_n(x):=\sqrt n \chi_{(0,n^{—1})}(x)$ on $[0,1]$; then $\int_{[0,1]}|f_n|^2dx=1$, and $f_n\to 0$ almost everywhere.

  4. It's true: apply Egoroff theorem and Cauchy-Bunyakovsky-Schwarz inequality. We can assume that $f=0$ (otherwise we consider $g_n:=f_n-f$ instead of $f_n$, which is integrable and $g_n\to 0$ pointwise). Fix $\varepsilon>0$. Then we can find $C$ measurable such that $\sup_{x\in C}|f_n(x)|\to 0$ and $\mu(X\setminus C)\lt \varepsilon$. We have $$\int_X|f_n(x)|d\mu(x)\leqslant \mu(X)\sup_{x\in C}|f_n(x)|+\sqrt{\varepsilon}\sqrt{\int_X |f_n|^2}d\mu,$$ so $$\limsup_{n\to+\infty}\int_X|f_n(x)|d\mu(x)\leqslant \sqrt \varepsilon\sup_{k\in\Bbb N}\sqrt{\int_X |f_k|^2}d\mu.$$

share|cite|improve this answer
Can you explain why we may assume $f=0$ in part 4? – analysis_qual May 1 '14 at 21:57
See edit. ${}{}{}$ – Davide Giraudo May 1 '14 at 22:16
I guess I'm having trouble seeing why replacing $f_n$ with $f_n - f$ preserves the assumption $\sup_n \| f_n \|_2^2 < \infty$. Is this obvious? – analysis_qual May 1 '14 at 22:41
It's not so obvious: we use (1). – Davide Giraudo May 2 '14 at 5:57

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.