# Assume that $f_n\to f$ in measure and $\sup_n\|f_n\|_{L^p(E)}<\infty$ for some $p>1$. Prove that $f_n$ converges to $f$ in $L^1$ norm.

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!

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.

• 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 $a_n \to 0$. – Nate Eldredge Aug 11 '15 at 4:41