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I am trying to do the following exercise in the Measure Theory book by Cohn

Prove that the Monotone Convergence Theorem still holds if the assumption that the functions $f_1, f_2, ...$ are non-negative is dropped, and the assumption that $f_1$ is integrable is added (note that in this case the integrals of the functions $f$ and $f_2, f_3, ...$ exist, but may equal $\infty$.)

I think I was able to solve it, but it took two cases, and case $2$ seems a bit contrived. I was wondering if my solution is correct, and if there is a better way? Thank you.

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Case 1: $f$ is integrable Since $f_1 \le f_2 \le \cdots$, the sequence $(f_n-f_1)_{n=0}^{\infty}$ is increasing, non-negative, and converges to $f-f_1$. Now we may apply the MCT to get $$\lim_{n \to \infty} \int (f_n-f_1) = \int(f-f_1)$$

Since each $f_n$ satisfies $f_1 \le f_n \le f$ and both $f_1$ and $f$ are integrable, each $f_n$ is integrable, so we may split the integral on the left. Since $f$ and $f_1$ is integrable, we may split the integral on the right. Cancelling, we get

$$\lim_{n \to \infty} \int f_n = \int f$$

Case 2: $f$ is not integrable Since $f_1$ is integrable, $\int(f_1)_- < \infty$, so $\int (f_n)_- < \infty$ and $\int (f)_- < \infty$. Therefore, since $f$ is not integrable, we must have that $\int f = \int f_+ = \infty$. Now apply the MCT to the sequence $(f_n)_+$ which converges to $f+$. We get

$$\lim_{n \to \infty} \int (f_n)_+ = \int (f_n)_+ = \infty$$

Now $$\lim _{n \to \infty} \int f_n = \lim _{n \to \infty} \left[ \int (f_n)_+ - \int (f_n)_- \right ]= \infty - \lim_{n \to \infty} \int (f_n)_- = \infty.$$

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The case analysis is important. There are other ways to do it, but they aren't really anymore elegant than how you do it.

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