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I am seeking solution verification/feedback for the following problem. Thank you very much in advance.

$\textbf{Problem:}$ Let the function $f: [0,1] \rightarrow \mathbb{R}$ be a non-negative, Lebesgue measurable function, and let $\lambda$ denote the Lebesgue measure on $[0,1]$. Prove that $f$ is Lebesgue integrable if and only if $\sum^{\infty}_{n=1} \lambda(\{x: f(x) \geq n\})$ converges.

$\textbf{Solution:}$ Assume $f$ is Lebesgue integrable. Then, $\int f \> d\lambda$ is finite. Put $E_n=\{x: f(x) \geq n\}$. We show $\sum^{\infty}_{n=1} \lambda(E_n)$ converges. For each $n$, $n\chi_{E_n} \leq f$. Hence, $$ n \int \chi_{E_n} \> d\lambda \leq \int f \> d\lambda$$ $$ n \lambda(E_n) \leq \int f \> d\lambda$$ $$\lambda(E_n) \leq \frac{1}{n} \int f \> d\lambda$$ Now, we see that $\lambda(E_n) \rightarrow 0$ as $n \rightarrow \infty$; as such, we conclude $\sum^{\infty}_{n=1} \lambda(E_n)$ converges.

$\>\>\>$Conversely, again put $E_n=\{x: f(x) \geq n\}$ and assume $\sum^{\infty}_{n=1} \lambda(E_n)$ converges. Define a sequence of functions $(\phi_n)$ by the following equation: $$\phi_n = \sum^{n}_{k=1} \chi_{E_k}.$$ This sequence is monotone and converges to $f$. The Monotone Convergence Theorem gives us that $$\int f \> d\lambda = \lim \int \phi_n \> d\lambda.$$ Since, $$\lim \int \phi_n \> d\lambda = \lim_{n} \sum^{n}_{k=1} \lambda(E_k) = \sum^{\infty}_{n=1} \lambda(E_n),$$ our assumption gives us that $\lim \int \phi_n \> d\lambda$ and, consequently, $\int f \> d\lambda$ is finite. $\blacksquare$

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  • $\begingroup$ Do you mean "f is integrable if and only if..."? Because you've already assumed $f$ to be measurable. $\endgroup$ – user61527 Aug 25 '13 at 21:15
  • $\begingroup$ Woah. Thank you! That was a major typo. :) $\endgroup$ – dgc1240 Aug 25 '13 at 21:23
  • $\begingroup$ It isn't sufficient for the terms to be positive and diminishing to zero for the series to converge, so I'm missing something in the first part. Another edit: no, I don't believe $\phi_n\to f$ pointwise, as its range is contained in $\mathbb{N}$ (no reason for $f$ to adhere to that condition). $\endgroup$ – Jonathan Y. Aug 25 '13 at 21:36
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    $\begingroup$ "Now, we see that $\lambda(E_n) \rightarrow 0$ as $n \rightarrow \infty$; as such, we conclude $\sum_n \lambda(E_n)$ converges." This is not true . For example $\frac 1 n \rightarrow 0$ but $\sum_n \frac 1n $ does not converge. $\endgroup$ – user42761 Aug 25 '13 at 22:10
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    $\begingroup$ Yes you can (but it's better to infer that $\limsup\sum_{n=1}^k\lambda(E_n)<\infty$, hence the positive series converges, as this approach doesn't assume the existence of a limit a priori. It's a moot point, seeing as how a positive series can't diverge except to infinity, but somehow it's more polite that way, to me at least). $\endgroup$ – Jonathan Y. Aug 26 '13 at 0:09
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My comment was probably not very illuminating; so here is a more detailed version. I apologize in advance because it does not really answer the question (dgc1240 was just asking for some feedback).

The idea is to use the well known formula $$\int_{[0,1]} f\, d\lambda=\int_0^\infty \lambda(\{ x;\; f(x)\geq t\})\, dt $$ which holds true for any nonnegative (measurable) $f$, possibly non-integrable.

This can be written as $$\int_{[0,1]}f\, d\lambda=\sum_{n=0}^\infty \int_n^{n+1}\lambda(\{ x;\; f(x)\geq t\})\, dt$$

Now, for each $n\geq 0$ we have $\lambda(E_{n+1})\leq \lambda(\{ x;\; f(x)\geq t\})\, dt\leq \lambda(E_n)$ whenever $t\in [n,n+1]$, so that $\lambda(E_{n+1})\leq \int_n^{n+1}\lambda(\{ x;\; f(x)\geq t\})\, dt\leq\lambda(E_n)$. It follows that $$\sum_{n=1}^\infty \lambda(E_n)\leq\int_{[0,1]} f\, d\lambda\leq \sum_{n=0}^\infty \lambda (E_n)\, , $$ so $f$ is integrable (i.e. $\int f\, d\lambda<\infty$) if and only if the series $\sum \lambda(E_n)$ is convergent.

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