Necessary and Sufficient Condition for Integrability in finite measure space

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Let (\Omega, \Sigma, \mu) be a finite measure space. Suppose that f\geq 0 is a measurable function on \Omega. Let E_n:=\{\omega\in\Omega:f(\omega)\geq n\} for each n\in \mathbb{N}\cup\{0\}. Show that f is integrable if and only if \sum_{n=0}^\infty \mu(E_n)<\infty.

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 g=\sum_{n=1}^\infty I_{E_n}.

Note that for each n\geq 0, n\in\mathbb{N} on E_n\setminus E_{n+1}, g=n, while n\leq f<n+1. Therefore g\leq f<g+1 on \Omega.

Integrating with respect to \mu, we get \int_{\Omega} gd\mu\leq\int_{\Omega} fd\mu<\int_{\Omega} g+1d\mu.

\displaystyle\boxed{\sum_{n=1}^\infty \mu (E_n)\leq \int_\Omega f d\mu<\sum_{n=1}^\infty\mu(E_n)+\mu(\Omega)}

(=>) Now assuming f is integrable, i.e. \int fd\mu<\infty, we have \sum_{n=1}^\infty \mu(E_n)<\infty. \mu(E_0)=\mu(\Omega)<\infty. Therefore \sum_{n=0}^\infty\mu(E_n)<\infty.

(<=) Conversely, if \sum_{n=0}^\infty\mu(E_n)=\sum_{n=1}^\infty \mu(E_n)+\mu(\Omega)<\infty, then \int_\Omega f d\mu<\infty.

We are done.

Note: For a more rigorous proof of \int gd\mu=\sum_{n=1}^\infty \mu (E_n) we can use MCT (Monotone Convergence Theorem).

Let g_k=\sum_{n=1}^k I_{E_n}. Then g_k\uparrow g. By MCT, \int gd\mu=\lim_{k\to\infty} \int g_k d\mu=\lim_{k\to\infty} \sum_{n=1}^k \mu (E_n)=\sum_{n=1}^\infty \mu(E_n).


About mathtuition88
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