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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?

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  • $\begingroup$ 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? $\endgroup$ Commented Jun 18, 2019 at 16:34

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$g$ 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 $$\int f = \sup_{\phi \le f} \int \phi \le 2 \int g < \infty.$$

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