I have the following question:
Let $X$: $\mu(X)<\infty$, and let $f \geq 0$ on $X$. Prove that $f$ is Lebesgue integrable on $X$ if and only if $\sum_{n=0}^{\infty}2^n \mu(\lbrace x \in X : f(x) \geq 2^n \rbrace) < \infty $.
I have the following ideas, but am a little unsure. For the forward direction:
By our hypothesis, we are taking $f$ to be Lebesgue integrable. Assume $\sum_{n=0}^{\infty}2^n \mu(\lbrace x \in X : f(x) \geq 2^n \rbrace) = \infty $. Then for any n, no matter how large, $\mu(\lbrace x \in X : f(x) \geq 2^n \rbrace)$ has positive measure. Otherwise, the sum will terminate for a certain $N$, giving us $\sum_{n=0}^{N}2^n \mu(\lbrace x \in X : f(x) \geq 2^n \rbrace) < \infty $. Thus we have $f$ unbounded on a set of positive measure, which in combination with $f(x) \geq 0$, gives us that $\int_E f(x) d\mu=\infty$. This is a contradiction to $f$ being Lebesgue integrable. So our summation must be finite.
For the reverse direction:
We have that $\sum_{n=0}^{N}2^n \mu(\lbrace x \in X : f(x) \geq 2^n \rbrace) < \infty \\$. Assume that $f$ is not Lebesgue integrable, then we have $\int_E f(x) d\mu=\infty$. Since we are integrating over a finite set $X$, then this means that $f(x)$ must be unbounded on a set of positive measure, which makes our summation infinite, a contradiction.
Any thoughts as to the validity of my proof? I feel as if there is an easier, direct way to do it.
Best Answer
$$\frac12\left(1+\sum_{n=0}^{+\infty}2^n\,\mathbf 1_{f\geqslant2^n}\right)\leqslant f\lt1+\sum_{n=0}^{+\infty}2^n\,\mathbf 1_{f\geqslant2^n}$$