Integrate the limit of a decreasing sequence of positive measurable functions in a finite measure space

Question

The question

Let $(X,\mathcal{A},\mu)$ be a finite measure space (i.e. $\mu(X)<\infty$), and let $f_n:X\to[0,\infty)$ be a (pointwise) decreasing sequence of measurable functions. Show that in this case the limit of the integrals is the integral of the limit:

$$\lim_{n\to\infty} \int f_n d\mu = \int \left(\lim_{n\to\infty} f_n\right) d\mu. $$

Note: from my research online, usually the hypothesis that $\mu$ is finite is replaced with the hypothesis that $f_1$ is integrable.

What I tried

As the question is speaking of strictly measurable functions and not necessarily integrable functions, my first intuition was to try to construct an increasing sequence of functions and apply the Monotone Convergence Theorem (MCT). The obvious thing to do is to let $g_n = f_1 - f_n$. Then $g_n$ forms an increasing sequence of positive measurable functions. The MCT directly gives that

$$\lim_{n\to\infty} \int (f_1 - f_n) d\mu = \int \left(f_1 - \lim_{n\to\infty} f_n \right)d\mu$$

and here I hit my first problem. For positive measurable functions $f$ and $g$, I know that $\int(f+g)d\mu = \int f d\mu + \int g d\mu$. Therefore, if $f\geq g\geq 0$ are measurable, then $\int f d\mu = \int (f-g)d\mu + \int g d\mu$. I would like to conclude that $\int fd\mu - \int gd\mu = \int (f-g)d\mu$, but for that I would need $g$ to be integrable, not merely measurable (I can't "substract infinity", the integral of $g$ needs to be finite). But supposing everything is fine, I get $$\int f_1d\mu - \lim_{n\to\infty}\int f_nd\mu = \int f_1d\mu - \int\left(\lim_{n\to\infty}f_n\right)d\mu$$ and the statement to be proven follows if the integral of $f_1$ is finite: proving this is my second problem.

I noticed that proving that $f_1$ is integrable would be strong enough to fix both my problems, because then each $f_n$ would be integrable by monotonicity, and so would be the integral of the limit of the $f_n$, by Fatou's Lemma. I also noticed that I didn't use the hypothesis of the finiteness of $\mu$. Am I on the right track? What more can I do?

Answer 1

The result is false even when $\mu (X) <\infty$. Let $f$ be non-negaive measurable function with $\int fd\mu=\infty$. Let $f_n=\frac f n$. Then $\int f_n d\mu =\infty$ for each $n$ but $\int \lim f_n=0$. In fact, finiteness of the measure plays no role at all.

If you assume that $f_1$ is integrable then $\int f_1 d\mu$ is, of course, finite and you already know how the proof goes in this case. If $\int f_1\mu=\infty$ then the result can fail: $f_n=\chi_{(n,\infty)}, f=0$ on the real line with Lebesgue measure is a counter-example.