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I hope you can help me with the question 10 of chapter 11 of the book Principles of Mathematical Analysis of Walter Rudin. My question is:

In this exercise, he asked to prove that if $\mu(X)<+\infty$ and $f\in L^2(\mu)$ on X, then $f\in L^1(\mu)$ on X. I thought strange to prove this proposition without the hypotheses that $f$ is mensurable. Do you think is possible to prove from $\mu(X)<+\infty$ and $f\in L^2(\mu)$ that $f$ is mensurable? Or the hypotheses that $f$ is mensurable is also necessary?

Thanks in advance!

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    It is good that you cite the source, but the title is misleading. It appears you ask the question itself, not something like "does $\mu(X)<+\infty$ and $f\in L^2(\mu)$ imply $f$ is measurable?" – 2017-01-03

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$L^2(\mu)$ contains only measurable functions, we have $$ L^2(\mu) = \left\{f \colon X \to \mathbf R : f \text{ is measurable}, \int_X |f|^2\, d\mu < \infty \right\} $$