# Question in the proof of Lebesgue's Dominated Convergence Theorem

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?

• "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!
– Did
Jul 8, 2017 at 9:23
• So the inequality is more like a assumption， I misunderstood it as an conclusion？@Did Jul 8, 2017 at 9:44
• 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 $$\lim_{n\to\infty}\int_X\left|\,f_n-\,f\right|\,d\mu=0$$
– Did
Jul 8, 2017 at 9:48

Use Fatou lemma and the fact that $2g-|f_{n}-f|\to 2g$ a.e as $f_{n}\to f$ a.e

• 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 $\le$ ? Jul 8, 2017 at 9:15
• You need to check the Fatou's Lemma! Check it once more.
– Riju
Jul 8, 2017 at 9:19
• 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? Jul 8, 2017 at 9:28
• But the inequality you have written comes from Fatou's Lemma. But aren't you asking about the inequality you have written down?
– Riju
Jul 8, 2017 at 9:37
• The stroke order in the book made me misunderstanding it as a precondition, that's the key of my question, thanks for your help. Jul 8, 2017 at 9:49

This is the Fatou Lemma,

https://en.wikipedia.org/wiki/Fatou%27s_lemma

Note that $\liminf_n (2 g - |f_n - f|)=2g$.

• But why the symbol of inequality is $\le$ not $\ge$ ? Jul 8, 2017 at 9:20
• 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.
– user178826
Jul 8, 2017 at 9:30