Monotone Decreasing Sequences of Functions: Is the Dominated Convergence Thm applicable?

Reading through Rudin's Real and Complex Analysis, I came across the following exercise:

Suppose $(f_n: X \to [0,\infty])$ is a monotone decreasing sequence of measurable functions such that $\lim\limits_{n \to \infty} f_n(x) = f(x)$ for all $x \in X$. Prove that if $f_1 \in L^1(\mu)$, then

$$\lim\limits_{n \to \infty}\int\limits_{X} f_n \, \mathrm{d}\mu = \int\limits_X f \, \mathrm{d}\mu.$$

It seems like this should be a trivial application of the Dominated Convergence Theorem, taking $f_1$ to be the dominating function. But it seems like an exercise would not be so trivial as to merit basically a one line proof. Is there a reason that DCT fails to be applicable here?

• I presume the exercise was meant to apply the monotone convergence theorem to $f_1-f_n$? Integrability just ensures that you can conclude $\int f_n \to \int f$ from $\int (f_1-f_n) \to \int (f_1-f)$. – copper.hat Jun 22 '13 at 17:57
• The exercise is listed after the DCT has been used somewhat at length, or I would think that the purpose was to motivate the theorem. Also taking the sequence $f_1-f_n$ is essentially replicating the general proof of the DCT. – Cameron Jun 22 '13 at 17:59
• DCT is certainly applicable. After all, $0 \leq f_n \leq f_1$ for all $n$. You should try to find a few examples showing that the result can fail if $f_1$ is not assumed to be integrable. – Martin Jun 22 '13 at 21:07