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Let $g$, $\{g_n\}$ be integrable and $f$, $\{f_n\}$ be measurable. Suppose $\mid{f_n}\mid\le{g_n}$ and $f_n\to f \ a.e$ and $\lim_{n \to \infty} \int{g_n}=\int g$. Show that $\lim_{n \to \infty} \int | f_n-f| =0$.

I will use dominated convergence theorem. But I can't find the dominating function.

zhw.
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asdwewqe
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  • You could have used a better title like "Variant of the dominated convergence theorem". – aexl May 15 '15 at 18:55
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    And I think you some more requirements, something like $g_n \to g$ a.e. – aexl May 15 '15 at 19:03
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    If you add the requirement $g_n \to g$ ae., then you can find an answer here http://math.stackexchange.com/q/72174/27978. – copper.hat May 15 '15 at 19:13

2 Answers2

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This is false (as @Alex S suggested). Working on $\mathbb {R}$ with Lebesgue measure, set $f_n=g_n = \chi_{[n,n+1]}.$ Then $f_n \to 0$ everywhere, and $\int g_n \to \int \chi_{[0,1]}.$ But clearly $f_n$ does not convverge to $0$ in $L^1.$

Let's add the hypothesis that $g_n \to g$ a.e. Then hint: Apply Fatou's Lemma to $g+g_n -|f-f_n|.$

zhw.
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Let $f_n = 1_{[n,n+1]}, f = 0$, $g_n = f_n$ and $g=1_{[0,1]}$, then we have $f_n(x) \to f(x)$ everywhere, $|f_n| \le g_n$, $\int g_n = \int g$, but $\int |f_n-f| = 1$ for all $n$.

copper.hat
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