# uniform $L^1(\mathbb{R})$ bound on a sequence implies it's a.e. limit has the same bound

Suppose $\{f_n\}\subset L^1(\mathbb{R})$ with $||f_n||_1\leq 1$ $\forall n$ and $f_n \to f$ a.e. How can I show that $||f||_1 \leq 1$? This will be easy once we know $f\in L^1(\mathbb{R})$ so I guess that is my question.

-
Use Fatou's Lemma. –  David Mitra Aug 6 '12 at 2:55
Here is an approach. Appealing to the inequality $|\,|a|-|b|\,| \geq | a | -|b|$ we have $$|\, ||f||_1 - ||f_n||_1 \,| \geq ||f||_1 - ||f_n||_1$$ $$\Rightarrow ||f||_1 - ||f_n||_1 \leq ||f||_1 - ||f_n||_1 \leq ||f_n||_1 - ||f||_1$$ $$\Rightarrow ||f||_1 \leq 2 ||f_n||_1 - ||f||_1 \Rightarrow ||f||_1 \leq ||f_n||_1 \leq 1$$ $$||f||_1 \leq 1$$ –  Mhenni Benghorbal Jan 25 at 23:07
@MhenniBenghorbal You do realize that $||a|-|b||\geq|a|-|b|$ does not imply that $|a|-|b|\leq |b|-|a|$? One can see the mistake without having any idea about measure theory or functional analysis. –  Michael Greinecker Jan 26 at 23:56
@MichaelGreinecker: Can you tell me what's wrong with this answer? It was just downvoted few minutes ago. Just to have an idea about what's going on and the attack on my answers. By the way, I have not said "I do not have wrong answers, offcourse, it is natural thing to make mistakes". Now, tell us what's wrong with the answer I referred you to. It is one of many which has been downvoted. –  Mhenni Benghorbal Jan 27 at 0:28
@MichaelGreinecker: How about these I, II, III. –  Mhenni Benghorbal Jan 27 at 0:35
You know that $|f_n|\to |f|$ pointwise a.e., this says that $\liminf |f_n|=|f|$ a.e. So in order to conclude what you want, by Fatou's Lemma, it's enough to show that $\liminf \int |f_n|\leq 1$.
Remember that: $$\liminf \int |f_n|=\sup\left\{\inf\left\{\int |f_n|,\int |f_{n+1}|,\int |f_{n+2}|,\ldots,\right\}:n\in \Bbb N\right\}.$$