# Very simple proof $f$ is integrable, small detail missing

Let $$f: \mathbb R \to \mathbb R$$ be measurable function, $$f \geq 0$$, $$f < \infty$$ almost everywhere. Define $$F_k = \{x \in \mathbb R: 2^k < f(x) \leq 2^{k+1}\}$$. We have that $$\displaystyle \{f(x) > 0\} = \text{supp} (f) = \bigcup_{k = -\infty}^{\infty}F_k$$ and $$F_k$$ are disjoint.

Show that $$f$$ is integrable if and only if $$\displaystyle \sum_{k=-\infty}^{\infty}2^km(F_k) < \infty$$.

What I did:

Define $$g(x) = \displaystyle \sum_{k=-\infty}^{\infty}2^k\chi_{F_k}(x)$$ and suppose $$f$$ is integrable. $$g$$ is clearly integrable as it is a "simple" measurable function.

Since $$g < f$$ and both are integrable we have $$\int g = \displaystyle \sum_{k=-\infty}^{\infty}2^km(F_k) < \int f < \infty$$

That was simple. Now suppose that $$\displaystyle \sum_{k=-\infty}^{\infty}2^km(F_k) < \infty$$.

We have the inequality $$g(x) < f \leq 2g(x)$$ from the construction of $$F_k$$ and $$g$$. Thus $$\int f \leq 2\int g < \infty$$ which concludes the proof.

Is this correct? How can we tell $$\int f$$ even makes sense? if a function is bounded between 2 other integrable functions, is it integrable?

• Your argument is incomplete. What is your definition of a "simple" measurable function? It is not simple in the usual sense of the word used in measure theory (i.e. a finite linear combination of indicator functions). Why is $g$ integrable? Jun 18, 2019 at 16:34

$$g$$ isn't simple but since $$g = \lim_{N \to \infty} \sum_{k=-N}^N 2^k \chi_{F_k}$$ is nonnegative you get $$\int g = \int \lim_{N \to \infty} \sum_{k=-N}^N 2^k \chi_{F_k} = \lim_{N \to \infty} \int \sum_{k=-N}^N 2^k \chi_{F_k} = \lim_{N \to \infty} \sum_{k=-N}^N 2^k m(F_k) = \sum_{k=-\infty}^\infty 2^k m(F_k)$$ as an application of the monotone convergence theorem.
Keep in mind that the integral of any nonnegative measurable function $$f$$ is defined: it just may happen to be infinite. Since you've shown that $$f \le 2g$$, any nonnegative simple function $$\phi \le f$$ also satisfies $$\phi \le 2g$$ so that $$\int \phi \le 2 \int g.$$ Thus $$\int f = \sup_{\phi \le f} \int \phi \le 2 \int g < \infty.$$