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I'm stuck with some problem of my Integral Calculation in Several Variables course. The problem goes like this:

Let $A\subset \mathbb{R}$ be a measurable set with $m(A)<\infty$, and $f:A\longrightarrow [0,\infty)$ a Lebesgue-measurable function. Prove that: $$f\in L^1(A)\Longleftrightarrow \sum_{n}^{\infty}m(\{ x\in A : f(x)\geq n \}) < \infty.$$

The notation I used is:

  • $m$ as the Lebesgue measure function
  • $L^1(A)=\{ f:A\rightarrow \mathbb{\overline{R}} : \int_{A}|f|\,\mathrm{d}m<+\infty \}$

I've started defining the set $A=f^{-1}([0,\infty))$ as a numerable sum of disjoint measurable sets (because it's said it's measurable) $\sum^{\infty}_{k=0}\cup I_k$, being each $I_k$ the real interval $[k,k+1)$. I imagine I should come to some conclusion like that any unbounded (upper bound) sets of $f(x)$ with $(x\in A)$ have measure $0$.

Xander Henderson
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Alejandro Bergasa Alonso
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    HINT: write the sets $f^{-1}([n,\infty ))$ as a disjoint union of sets $f^{-1}([k,k+1))$ and after use Tonelli's theorem in the series – Masacroso May 25 '20 at 09:30
  • @mathworker21 I know how to use Tonelli. I was struggling with the final step of the first part and the first step of the second one, the ones that say that $$\int\sum_{k\geqslant 0}(k+1)\mathbf{1}{f^{-1}([k,k+1)} \mathop{}!dm \geqslant \int f\mathop{}!dm$$ and $$\int f \mathop{}!dm \geqslant \int\sum{k\geqslant 0}k ,\mathbf{1}_{f^{-1}([k,k+1)}\mathop{}!dm.$$ The rest is clear to me, but I don't know if that inequalities are trivial and can't figure it out by myself, so I started the bounty to draw attention for somebody to clear that step up. – Alejandro Bergasa Alonso Jun 03 '20 at 19:02
  • Ok finally I've just figured it out. I'll give the bounty prive to the answer below, he posted a very good and elegant answer and deserves it. – Alejandro Bergasa Alonso Jun 03 '20 at 19:08
  • @AlejandroBergasaAlonso So in the description of the bounty, say "I need someone to explain the two uses of Tonelli in the answer below" rather than "I need a well explained step-by-step solution". – mathworker21 Jun 03 '20 at 19:13
  • Ok, I'll try to be more specific the next time. Although, my problem was not about Tonelli, but about the two inequalities i commented above, but i've already figured them out. – Alejandro Bergasa Alonso Jun 03 '20 at 19:20
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    Tonelli's theorem says that if you have two $\sigma $-finite measures $\mu$ and $\nu $ and a non-negative integrand $f$ measurable respect to the product measure $\mu \times \nu $ then $$\int f \mathop{}!d (\mu \times \nu )=\iint f\mathop{}!d \mu \mathop{}!d \nu =\iint f\mathop{}!d \nu \mathop{}!d \mu $$ In this question the two $\sigma $-finite measures are the counting measure (the infinite sum) and the Lebesgue measure. In short Tonelli's theorem assert that you can exchange the order of integration respect two $\sigma $-finite measures when the integrand is non-positive – Masacroso Jun 03 '20 at 19:20
  • @Masacroso Thanks! I guess the fact of $\int f dm$ being finite is due to sandwich theorem, because it's bounded by two integrals that, being $\sum_{n=0}^{\infty}m({x\in A : f(x)\geq n}) < \infty$, then are finite too. Am I correct? – Alejandro Bergasa Alonso Jun 03 '20 at 19:25
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    Alternatively to Tonelli's theorem you can exchange the sum and the integral using the monotone convergence theorem, because you are adding non-negative elements, so there is a sequence of functions that increases. Take a look here for a brief explanation of both methods. – Masacroso Jun 03 '20 at 19:25
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    Yes, the integral is sandwiched by two series that are almost the same series except a finite addend. Then if one of the series diverges the other diverges too so the value of the integral is infinite and viceversa, if some of the series converge the other too so the value of the integral is finite. – Masacroso Jun 03 '20 at 19:29

1 Answers1

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As stated in the comment you have that $$ f^{-1}([n,\infty ))= \bigcup_{k\geqslant n}f^{-1}([k,k+1)) $$ Therefore $$ \begin{align*} \sum_{n\geqslant 0}m(f^{-1}([n,\infty ))&= \sum_{n\geqslant 0}m\left( \bigcup_{k\geqslant n}f^{-1}([k,k+1))\right)\\ &= \sum_{n\geqslant 0}\sum_{k\geqslant n}m(f^{-1}([k,k+1))\\ &\overset{(*)}{=} \sum_{k\geqslant 0}\sum_{0\leqslant n\leqslant k}m(f^{-1}([k,k+1))\\ &=\sum_{k\geqslant 0}(k+1)m(f^{-1}([k,k+1))\\ &= \sum_{k\geqslant 0}\int(k+1)\mathbf{1}_{f^{-1}([k,k+1)} \mathop{}\!dm\\ &\overset{(**)}{=} \int\sum_{k\geqslant 0}(k+1)\mathbf{1}_{f^{-1}([k,k+1)} \mathop{}\!dm\\ &\geqslant \int f\mathop{}\!dm \end{align*} $$ where we used Tonelli's (or the monotone convergence) theorem in $(*)$ and $(**)$. Similarly $$ \begin{align*} \int f \mathop{}\!dm &\geqslant \int\sum_{k\geqslant 0}k \,\mathbf{1}_{f^{-1}([k,k+1)}\mathop{}\!dm\\ &=\sum_{k\geqslant 0}k\, m(f^{-1}([k,k+1))\\ &=\sum_{k\geqslant 1}\sum_{1\leqslant n\leqslant k} m(f^{-1}([k,k+1))\\ &= \sum_{n\geqslant 1} \sum_{k\geqslant n}m(f^{-1}([k,k+1))\\ &= \sum_{n\geqslant 1} m(f^{-1}([n,\infty )) \end{align*} $$ Finally note that $m(f^{-1}[0,\infty ))= m(A)<\infty $, so the answer is clear.

Masacroso
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  • Thank you for your answer! What does your notation $\mathbf{1}_{f^{-1}([k,k+1)}$ mean? – Alejandro Bergasa Alonso May 25 '20 at 18:02
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    it is an indicator function, that is, $\mathbf{1}{A}(x)=1$ if and only if $x\in A$, and $\mathbf{1}{A}(x)=0$ when $x\notin A$ – Masacroso May 25 '20 at 18:09
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    @AlejandroBergasaAlonso to understand the inequality of the series just note that $k\leqslant f, \mathbf{1}{f^{-1}[k,k+1)}\leqslant k+1$, that is, when we multiply $f$ by $\mathbf{1}{f^{-1}[k,k+1)}$ we are restricting $f$ to the subset where it image is contained in $[k,k+1)$, so the minimum value of $f$ in this set is $k$ or greater, and it supremum is at most $k+1$ – Masacroso Jun 04 '20 at 02:17
  • Perfect, I understand it. I'm struggling proving the simultaneous convergence of all the three expressions, and prove separately that the ones with sums converge. – Alejandro Bergasa Alonso Jun 04 '20 at 06:33