# Proving slight extension of the Monotone Convergence Theorem

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