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Here is a proof of the dominated convergence theorem.

Theorem. Suppose that $f_n$ are measurable real-valued functions and $f_n(x) \to f(x)$ for each $x$. Suppose there exists a nonnegative integrable function $g$ such that $|f_n(x)| \le g(x)$ for all $x$. Then$$\lim_{n \to \infty} \int f_n\,d\mu = \int f\,d\mu.$$

Proof. Since $f_n + g \ge 0$, by Fatou's lemma,$$\int f + \int g = \int (f + g) \le \liminf_{n \to \infty} \int (f_n + g) = \liminf_{n \to \infty} \int f_n + \int g.$$Since $g$ is integrable,$$\int f \le \liminf_{n \to \infty} \int f_n.\tag*{$(*)$}$$Similarly, $g - f_n \ge 0$, so$$\int g - \int f = \int (g - f) \le \liminf_{n \to \infty} \int (g - f_n) = \int g + \liminf_{n \to \infty} \int (-f_n),$$and hence$$-\int f \le \liminf_{n \to \infty} \int (-f_n) = -\limsup_{n \to \infty} \int f_n.$$Therefore$$\int f \ge \limsup_{n \to \infty} \int f_n,$$which with $(*)$ proves the theorem.$$\tag*{$\square$}$$

My question is as follows. Can we get another proof of the dominated convergence theorem by applying Fatou's lemma to $2g - |f_n - f|$?

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  • $\begingroup$ Hi I have a very naive question, why can you split the lim inf in the first line of the proof?? I understand that, at least for sequences of number, you have that: lim inf (a_n+b_n) >= lim inf a_n + lim inf b_n. thanks $\endgroup$ – Cristian Baeza Oct 6 '17 at 18:19
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Yes, absolutely. And actually, applying Fatou to $2g - \lvert f_n - f \rvert$ gives the stronger result that $$\int \lvert f_n -f \rvert d \mu \to 0$$ as $n\to \infty$. From this and $$\left \lvert \int f_n d\mu - \int f d\mu \right \rvert = \left \lvert \int (f_n - f) d\mu \right \rvert \le \int \lvert f_n -f \rvert d\mu$$ we recover the slightly weaker version that is proven above. The dominated convergence theorem is ordinarily proven using $2g - \lvert f_n - f \rvert$.

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