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Let be a finite measure space. Suppose that is a measurable function on . Let for each . Show that is integrable if and only if .

This proof has a cute solution that is potentially very short. We will elaborate more on this proof. Other approaches include using Markov’s Inequality / Chebyshev’s inequality.

Proof: Consider .

Note that for each on , , while . Therefore on .

Integrating with respect to , we get .

(=>) Now assuming f is integrable, i.e. , we have . . Therefore .

(<=) Conversely, if , then .

We are done.

Note: For a more rigorous proof of we can use MCT (Monotone Convergence Theorem).

Let . Then . By MCT, .

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