## 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. Therefore $g\leq f 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, $\int gd\mu=\lim_{k\to\infty} \int g_k d\mu=\lim_{k\to\infty} \sum_{n=1}^k \mu (E_n)=\sum_{n=1}^\infty \mu(E_n)$.