In the book "Real and Complex analysis 3rd ed." by Walter Rudin,in the proof of Lebesgue's Dominated Convergence Theorem (Chapter 1, page 26), since $\left|f\right|\le g,$ and $f$ is measurable, $f\in L^1(\mu)$. since $\left|f_n-f\right|\le 2g $, then: $$\int_X{2g\,d\mu}\leq\lim_{n\to\infty}\inf\int_X{2g-\left|\,f_n-\,f\right|\,d\mu}$$ but $\left|\,f_n-\,f\right|\ge0$, why could the inequality holds?
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1$\begingroup$ "why could the inequality hold(s)?" The inequality can only hold if $$\liminf_{n\to\infty}\int_X-\left|\,f_n-\,f\right|\,d\mu\geqslant0$$ which is equivalent to $$\limsup_{n\to\infty}\int_X\left|\,f_n-\,f\right|\,d\mu\leqslant0$$ which is equivalent to $$\lim_{n\to\infty}\int_X\left|\,f_n-\,f\right|\,d\mu=0$$ which is what is to be proven. Thus: Mission accomplished! $\endgroup$– DidJul 8, 2017 at 9:23
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$\begingroup$ So the inequality is more like a assumption， I misunderstood it as an conclusion？@Did $\endgroup$– NFDreamJul 8, 2017 at 9:44
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$\begingroup$ No, rather, the inequality $$\int_X{2g\,d\mu}\leq\liminf_{n\to\infty}\int_X(2g-\left|\,f_n-\,f\right|)\,d\mu$$ is proven (hence it is a partial conclusion) and then one notes that it implies (actually, is equivalent to) the desired result that $$\lim_{n\to\infty}\int_X\left|\,f_n-\,f\right|\,d\mu=0$$ $\endgroup$– DidJul 8, 2017 at 9:48
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2 Answers
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Use Fatou lemma and the fact that $2g-|f_{n}-f|\to 2g$ a.e as $f_{n}\to f$ a.e
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$\begingroup$ But the inequality must be $\int_X{2g\,d\mu}\ge\lim_{n\to\infty}\inf\int_X{2g-\left|\,f_n-\,f\right|\,d\mu}$. why it's $\le$ ? $\endgroup$– NFDreamJul 8, 2017 at 9:15
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$\begingroup$ You need to check the Fatou's Lemma! Check it once more. $\endgroup$– RijuJul 8, 2017 at 9:19
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$\begingroup$ I think it does not reach the step of Fatou's Lemma, of course that's the next step of the proof , but maybe not my question? $\endgroup$– NFDreamJul 8, 2017 at 9:28
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$\begingroup$ But the inequality you have written comes from Fatou's Lemma. But aren't you asking about the inequality you have written down? $\endgroup$– RijuJul 8, 2017 at 9:37
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$\begingroup$ The stroke order in the book made me misunderstanding it as a precondition, that's the key of my question, thanks for your help. $\endgroup$– NFDreamJul 8, 2017 at 9:49
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This is the Fatou Lemma,
https://en.wikipedia.org/wiki/Fatou%27s_lemma
Note that $\liminf_n (2 g - |f_n - f|)=2g$.
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$\begingroup$ But why the symbol of inequality is $\le$ not $\ge$ ? $\endgroup$– NFDreamJul 8, 2017 at 9:20
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1$\begingroup$ For sure you have $\geq$, because $2 g - |f-f_n|\leq 2 g$. But, if you open the link I wrote you, you can see that the Fatou's Lemma gives you the other inequality. Both things together mean that you have even the equality. $\endgroup$– user178826Jul 8, 2017 at 9:30