# 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.

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Use Fatou's Lemma. –  David Mitra Aug 6 '12 at 2:55
@DavidMitra Your can post your comment as an answer. –  Davide Giraudo Aug 10 '12 at 12:33

Just following the David Mitra's hint, this is the Fatou's Lemma from Zygmund & Whedeen Measure and Integral:

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: $$\liminf \int |f_n|=\sup\left\{\inf\left\{\int |f_n|,\int |f_{n+1}|,\int |f_{n+2}|,\ldots,\right\}:n\in \Bbb N\right\}.$$

Can you catch it from here?

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Appealing to the inequality $|\,|a|-|b|\,| \geq | a | -|b|$ we have $$|\, ||f||_1 - ||f_n||_1 \,| \geq ||f||_1 - ||f_n||_1$$ $$\Rightarrow ||f||_1 - ||f_n||_1 \leq ||f||_1 - ||f_n||_1 \leq ||f_n||_1 - ||f||_1$$ $$\Rightarrow ||f||_1 \leq 2 ||f_n||_1 - ||f||_1 \Rightarrow ||f||_1 \leq ||f_n||_1 \leq 1$$ $$||f||_1 \leq 1$$
This reproduces (and is as wrong as) your first solution, now deleted (which two users commented by (1.) pointing at the problem and (2.) giving a counterexample). Exercise: find $(f_n)$ like in the question such that $\|f\|_1\gt\|f_n\|_1$ for every $n$ (thus you try to show an intermediate result which is false). –  Did Aug 17 '12 at 7:26