If $\{f_n\}$ is a sequence of measurable functions on the measurable set $E$ such that $f_n \rightarrow f$ converges pointwise almost everywhere on $E$ where $|f| \in L(E)$ "i.e. $|f|$ is Lebesgue integrable on $E$, say, $\int_{E} |f| = a$ and $\lim_{n \rightarrow \infty} \int_{E} |f_n(x)|= b$.

I need to show that $\lim_{n \rightarrow \infty} \int_{E} |f_n(x)-f(x)|$ exists and to find its value. I can see that $|f_n(x)-f(x)| \leq |f_n(x)|+|f(x)|$ so passing the integral over $E$ then taking the limit as $n \rightarrow \infty$ proves that the limit of the integral exists, but I am still not sure how to find the limit value, my guess is to apply the Lebesgue Dominated Convergence Theorem or the general Lebesgue Dominated Convergence Theorem.

  • $\begingroup$ Hmm. $(-1)^n\le 2$ and $\lim 2$ exists, hence $\lim (-1)^n$ exists? $\endgroup$ – David C. Ullrich Mar 23 at 21:43
  • $\begingroup$ I passed the limit into positive terms, right? $\endgroup$ – math_for_ever Mar 23 at 22:30
  • $\begingroup$ so $0<2+(-1)^n \le 4$ and $\lim 4$ exists, hence $\lim(2+(-1)^n)$ exists. got it, I didn't see how positivity mattered... $\endgroup$ – David C. Ullrich Mar 23 at 22:57

Here is a hint: apply Fatou's lemma to $|f_n - f| + |f| - |f_n| \ge 0$ to find that $$b-a \le \liminf \int |f_n - f|.$$ Then apply Fatou's lemma to $|f_n| + |f| - |f_n - f|$ to find in turn that $$\limsup \int |f_n - f| \le b-a.$$

  • $\begingroup$ By the way this forces $a \le b$. $\endgroup$ – Umberto P. Mar 23 at 21:53
  • $\begingroup$ Thank you ! that makes the problem more sense. $\endgroup$ – math_for_ever Mar 23 at 22:28

It seems like this, I am not sure if it is correct.

Please tell me where I was wrong.

Since $\lim_{n}\int_E|f_n(x)|=b \lt ∞ $, $ {\exists}$ $N$ such that ${\forall} n \gt N$$f_n\in L(E)$.

Since $f_n \rightarrow f a.e.$, we have $f_n - f \rightarrow 0 a.e.$.

And $|f_n - f|$ domined by $|f_n| + |f|$, by Lebesgue Dominated Convergence Theorem $\int_E|f_n - f| \rightarrow \int_E 0 =0$.


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