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a homework question from measure and integraiton theory course.

Suppose $f_n \in L_1(\mathbb R^d)$ for each $n\in\mathbb N$, $f_n\geq 0$ , $f_n\to f$ a.e and $\int f_n\to\int f<\infty$.
Prove that $\int|f_n-f| \to 0$

(Hint: ($f_n - f)_-\leq f$. Use dominated convergence theorem )

I am thinking that since $|f_n -f | \to 0$ a.e and $|f_n-f|=(f_n - f)_+ + (fn-f)_-$ . If I can show $(f_n-f)_+ ≤ f$ and $(f_n-f)_-≤f$ . Then since $f_n$ and $f$ are integrable , by DCT, $\int|f_n-f| \to \int0 =0$ . I think my approach might be wrong ..

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up vote 1 down vote accepted

Thomas E.'s approach is fine, but you can get this with a different use of the triangle inequality somewhat easier using the inequality $0 \leq |a - b| + |a| - |b| \leq 2|a|$. The proof of this is straightforward: We have $|a-b|+|a|-|b|\leq2|a|$ since $|a-b|\leq|a|+|b|$. Further, $||a|-|b||\leq|a-b|$ so that if $|a|\leq|b|$ then $0\leq|a-b|+|a|-|b|$. Similarly if $|b|\leq|a|$ then we have $0 \leq |a-b| + |a|-|b|$.

Apply the dominated convergence theorem to $0 \leq |f_n - f| + f - f_n \leq 2f$ (since in your case everything is positive) and the result follows.

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Going to positive and negative parts is not really necessary. Here's another type of approach that is also quite straight-forward.

Denote $g_{n}=|f|+|f_{n}|-|f-f_{n}|$ for all $n$, which are non-negative (by triangle-inequality) and measurable. Apply Fatou's Lemma and use the fact that $\liminf(-a_{n})=-\limsup(a_{n})$.

If I calculated them correctly, then what you should get is $$\liminf_{n\to\infty}\int g_{n}=\|f\|-\limsup_{n\to\infty} \|f-f_{n}\|$$ and $$\int \liminf_{n\to\infty} g_{n} =\|f\|,$$

where $\|\cdot\|$ is the $L^{1}$-norm. Do you see how this implies your result?

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You can use \| for norm, as it has a correct spacing. In large quantities it even improves readability by quite a lot. – Asaf Karagila May 25 '12 at 15:23
Sure, didn't know that. Thanks. – T. Eskin May 25 '12 at 15:24

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