# Alternate proof of the dominated convergence theorem by applying Fatou's lemma to $2g - |f_n - f|$?

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 $$2g - |f_n - f|$$?

• 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 Oct 6 '17 at 18:19

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 $2g - \lvert f_n - f \rvert$.