Let $\{f_n\}$ and $f$ be Lebesgue measurable functions on $E$ where $|E|<\infty$. Assume that $f_n\to f$ in measure and $\sup_n\|f_n\|_{L^p(E)}<\infty$ for some $p>1$.

(a) Prove that $f_n$ converges to $f$ in $L^1$ norm.

(b) Show by counterexample that this convergence may no longer hold if you replace the $L^p$ condition with $\sup_n \|f_n\|_{L^1(E)}$.

I can deal with the show by counterexample part. Let $f_n(x)=n\chi_{[0,\frac{1}{n}]}$ and $E=[0,1]$, then $f_n(x)\to 0$ in measure. However, $\int_0^1 f_n(x)-f(x)dx=1$.

I am stuck in proving part (a), I tried to use Egorov theorem, which requires $f_n\to f$ a.e. Since $f_n \to f$ in measure, then there is a subsequence $f_{n_k}$ converge a.e. Then $f_{n_k}$ converges uniformly in a compact set $F$. So $$\int_{E}|f_{n_k}(x)-f(x)|dx=\int_{F}|f_{n_k}(x)-f(x)|dx+\int_{E\backslash F}|f_{n_k}(x)-f(x)|dx$$ By the uniformly convergence of $f_{n_k}$ on $F$, then $\int_{F}|f_{n_k}(x)-f(x)|dx<\epsilon M$.

I have trouble with the $\int_{E\backslash F}|f_{n_k}(x)-f(x)|dx$ part.

$\int_{E\backslash F}|f_{n_k}(x)-f(x)|dx<\|f_{n_k}(x)-f(x)\|_{L^p(E\backslash F)}\|\Bbb{1}\|_{L^q(E\backslash F)}<\epsilon \|f_{n_k}(x)-f(x)\|_{L^p(E\backslash F)} $ by Holder's inequality.

Then how to show $\|f_{n_k}(x)-f(x)\|_{L^p(E\backslash F)}$ is actually bounded by the condition given in the problem? And if $\|f_{n_k}(x)-f(x)\|_{L^p(E\backslash F)}$ is bounded, it only proves the subsequence $f_{n_k}$ converges in $L^1$ norm, how to show the $f_n$ also converges in $L^1$ norm?

Could someone kindly help? Thanks!


1 Answer 1


You've done the bulk of the work already. To show that $\|f_{n_k} - f \|_{L^p(E \setminus F)}$ is bounded, note that by the triangle (Minkowski's) inequality, $$\|f_{n_k}-f \|_{L^p(E \setminus f)} \leq \|f_{n_k}\|_{L^p(E)} + \|f\|_{L^p(E)}.$$ By Fatou's lemma, $$\|f\|_{L^p(E)}^p = \int_E |f|^p = \int_E \lim\limits_{k \to \infty} |f_{n_k}|^p \leq \liminf\limits_{k \to \infty} \int |f_{n_k}|^p \leq \sup\limits_n \|f_n\|^p < \infty.$$

To take care of the subsequence issue, the following applies in a more general setting, and is quite useful. By what is already shown, we have that any subsequence of $f_n$ has a sub-subsequence that converges to $f$. Thus, the set $\{f_n\}_n$ has compact closure in $L^1$. The following is true: let $(X,d)$ be a compact metric space, and let $\{x_n\}_{n=1}^\infty$ be a sequence in $X$ such that every convergent subsequence of $\{x_n\}$ converges to the same limit $x$. Then $\{x_n\}$ converges to $x$. Proof: suppose not. Then there exists $\epsilon>0$ and a subsequence $\{x_{n_k}\}_{k=1}^\infty$ with $d(x_{n_k},x)>\epsilon$ for all $k$. Since $X$ is compact, $\{x_{n_k}\}$ has a convergent subsequence $\{x_{n_{k_j}}\}_{j=1}^\infty$. But $\{x_{n_{k_j}}\}$ is a convergent subsequence of $\{x_n\}$, and therefore must converge to $x$, contradicting our definition of $\{x_{n_k}\}$. Apply this to the closure of your sequence in $L^1$ to complete the proof.

  • $\begingroup$ Alternatively, note that if $M = \sup_n \|f_n\|_{L^p(E)}$, then by Holder we have $\|f_n\|_{L^1(E)} \le M |E|^{1/q}$. Fatou says we also have $\|f\|_{L^1(E)} \le M |E|^{1/q}$. So if we set $a_n = \|f_n -f\|_{L^1(E)}$, the triangle inequality gives $0 \le a_n \le 2 M |E|^{1/q}$. Now apply the double subsequence lemma described above in the compact metric space $[0, 2M|E|^{1/q}]$, and conclude that $a_n \to 0$. $\endgroup$ Commented Aug 11, 2015 at 4:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .