Let $f: X\to\mathbb{R}$ be bounded and measurable (X meas.). Show that $f$ is Lebesgue integrable

Question

Let $f: X\to\mathbb{R}$ be bounded and measurable and $X$ is a measurable subset of $\mathbb{R}$. Suppose $\exists M>0$, $\alpha \in (0,1)$, s.t $\forall \epsilon > 0$

$$ m(\{x \in X: |f(x)| > \epsilon \}) < \frac{M}{\epsilon^{\alpha}} $$

Show that $f$ is Lebesgue integrable.

I already solve this problem when $f: [0,1] \to \mathbb{R}$ and when M is not in the problem. That is easier, because only considering a simple partition of [0,1] we can get that $\int|f(x)| < \infty$. But on this problem, I have a measurable set $X$ and it's not given that it has finite measure and that complicates the problem. I believe I have to use the simple approximation theorem.

Answer 1

Hint: By Fubini's theorem \begin{align*}\int_X |f(x)|m(dx) &= \int_X \int_0^{\infty} 1_{\{t \leq |f(x)|\}} dt\; m(dx) \\ &= \int_0^{\infty} \int_X 1_{\{t \leq |f(x)|\}} m(dx) \;dt \\ & = \int_0^{\infty} m(\{x:|f(x)|\geq t\})dt \\ &= \int_0^Bm(\{x:|f(x)| \geq t\})dt\end{align*} where $B$ is the upper bound for $|f|$. Now use the given condition to show that this integral is finite.

Answer 2

Suppose $f$ is bounded by $C$. Set $A = \{x\in X : f(x) \neq 0\}$ and $A_n = \{x\in A : C\cdot 2^{-n-1} \le \lvert f(x)\rvert \le C\cdot 2^{-n}\}$. Note $\int_X \lvert f\rvert\, dm = \int_A \lvert f\rvert\, dm$. As $A$ is the disjoint union of the $A_n$, the monotone convergence theorem gives $$\int_A \lvert f\rvert\, dm = \sum_{n\, =\, 0}^\infty \int_{A_n} \lvert f\rvert\, dm$$ Using the hypothesis, the latter sum is dominated by the geometric series

$$2^\alpha C^{1-\alpha}M\sum_{n\, =\, 0}^\infty 2^{-(1-\alpha)n}$$

which converges since $1-\alpha > 0$.