# analysis-integration measure

Given a measurable space $(X, V, m)$ and $\{F_{n}\}_{1}^{\infty}\subset$ $V$ is a sequence of sets such that $m(F_{n})\leq$ $e^{-n}$ $\forall {n}.$ show that the functions $h(x)=\sum_{1}^{\infty} { \chi_E{_n}(x)}$ and $g(x)=\sum_{1}^{\infty} {n^{t}\chi_E{_n}(x)}$ belongs to $L^p$ for all $0<p, t<\infty$.

-
Is $E^n = F_n$? –  Rudy the Reindeer Nov 12 '11 at 21:46
Instead of deleting your old question, you could have edited the old one. This would have bumped it to the front page. At that opportunity you could also have gotten rid of the typos that were already pointed out to you. –  t.b. Nov 12 '11 at 21:58
@ t.d, I wanted to make use of the definition of L^p, that was why I deleted the other equations. –  wright Nov 12 '11 at 22:18
You should think about accepting an answer. =) –  Patrick Da Silva Nov 13 '11 at 19:43

Since $$\left( \int_X \left( \sum_{n=1}^N \chi_{F_n} \right)^p dm \right)^{1/p} \le \sum_{n=1}^N \left( \int_X \chi_{F_n} dm \right)^{1/p}$$ (i.e. $\|f+g\|_p \le \|f\|_p + \|g\|_p$ and by induction, the same goes for $n$ functions), by letting $N$ go to infinity, we can use the monotone convergence theorem because since $0 \le \sum_{n=1}^N \chi_{F_n} \le \sum_{n=1}^{N+1} \chi_{F_n}$, this sequence is increasing, thus $$\| h \|_p = \left\| \lim_{N \to \infty} \sum_{n=1}^N \chi_{F_n} \right\|_p = \lim_{N \to \infty} \left\| \sum_{n=1}^N \chi_{F_n} \right\|_p \le \lim_{N\to \infty} \sum_{n=1}^N \| \chi_{F_n} \|_p = \sum_{n=1}^{\infty} \|\chi_{F_n} \|_p$$ (for the second equality, putting both sides at $p^{\text{th}}$ power is like looking at integrals, thus I can use monotone convergence). Now using the properties of the $F_n$'s we obtain \begin{align*} \|h\|_p & \le \sum_{n=1}^{\infty} \left( \int_X \chi_{F_n} \, dm \right)^{1/p} = \sum_{n=1}^{\infty} \left( m(F_n) \right)^{1/p} \\ & \le \sum_{n=1}^{\infty} (e^{-n})^{1/p} = \sum_{n=1}^{\infty} (e^{-1/p})^n = \frac{e^{-1/p}}{1-e^{-1/p}} < \infty. \end{align*} (I assumed your $E_n$'s were the $F_n$'s.)

Some similar argument should hold for the next one, I just gave you the ideas so that you can still try the next. If you want me to work it out the second one I will, just comment.

Hope that helps,

EDIT : You seem to have trouble with the second case. Notice that the same steps go for $g_t$ up to some point, i.e. $$\|g_t \|_p \le \sum_{n=1}^{\infty} n^t (e^{-1/p})^n.$$ Now the series $\sum_{n=1}^{\infty} n^t x^n$ converges when $|x|<1$ for all $t \in \mathbb R$. If $t \le 0$ this is trivial (the sum is bounded by the geometric series, thus converges), so let me suppose that $t > 0$. If $t$ is not an integer, then the sum computed at $t$ is less than the sum computed at an integer greater than $t$, so I'll assume $t$ is an integer. Then $$\sum_{n=1}^{\infty} n^t |x|^n \le \sum_{n=1}^{\infty} (n+1)(n+2)\cdots(n+t)|x|^n$$ and the sum on the right is the $t^{\text{th}}$ derivative of the geometric series, thus it converges absolutely with the same radius of convergence.

-
Yea, it helps but I think there was a little typo in your last equation. That is a geometric series but I understand your statements. –  wright Nov 12 '11 at 22:29
Uh, yeah my last line was a little weird. Let me correct that. –  Patrick Da Silva Nov 12 '11 at 22:46
Ok. I have worked on the second but in this case, you don't end up getting a geometric series and I think I can its less than infinty or? –  wright Nov 12 '11 at 22:55
After thinking a little about it, I didn't see a typo... where did you? I edited my answer because I noticed I didn't care much about the limit statements and trusted the nice behavior of $L^p$ norms a little too much... –  Patrick Da Silva Nov 12 '11 at 23:00
Use the fact that $\sum_{n=1}^{\infty} n^t x^n < \infty$ when $|x|< 1$ and $t \in \mathbb R$. –  Patrick Da Silva Nov 12 '11 at 23:04

Assuming $E^n = F_n$.

To show $h \in L^p$ you need to show $(\int_X |h|^p )^{1/p} < \infty$:

$$\left( \int_X |h|^p \right)^{1/p} = \left( \int_X |\sum \chi_{F_n}|^p \right)^{1/p} \leq \left( \int_X \left( \sum |\chi_{F_n}| \right)^p \right)^{1/p}$$

Then using the Lebesgue dominated convergence theorem you know that you can do this:

$$\left( \int_X \left( \sum |\chi_{F_n}| \right)^p \right)^{1/p} = \left( \int_X \lim_{N \rightarrow \infty} \left( \sum_{n=1}^N |\chi_{F_n}| \right)^p \right)^{1/p} = \left(\lim_{N \rightarrow \infty} \int_X \left(\sum_{n=1}^N |\chi_{F_n}| \right)^p \right)^{1/p}$$

i.e. you can swap limit and integral and then because you have a finite sum you can swap integral and sum. Let's put things together:

\begin{align*} \left( \int_X |h|^p \right)^{1/p} = \left( \int_X \Big |\sum_{n=1}^\infty \chi_{F_n} \Big |^p dm \right)^{1/p} \\ \stackrel{\Delta-ineq.}{\leq} \left( \int_X \left( \sum_{n=1}^\infty |\chi_{F_n}| \right)^p dm \right)^{1/p} \\ = \left( \int_X \left( \lim_{N \rightarrow \infty} \sum_{n=1}^N |\chi_{F_n}| \right)^p dm \right)^{1/p} \\ \stackrel{cont. of ()^p}{=} \left( \int_X \lim_{N \rightarrow \infty} \left( \sum_{n=1}^N |\chi_{F_n}| \right)^p dm \right)^{1/p} \\ \stackrel{Lebesgue}{=} \left( \lim_{N \rightarrow \infty} \int_X \left( \sum_{n=1}^N |\chi_{F_n}| \right)^p dm \right)^{1/p} \\ \stackrel{\|.\| \Delta ineq.}{\leq} \lim_{N \rightarrow \infty} \sum_{n=1}^N \left( \int_X |\chi_{F_n}|^p dm \right)^{1/p} \\ \stackrel{\chi \in \{0,1\}}{=}\lim_{N \rightarrow \infty} \sum_{n=1}^N \left( \int_X \chi_{F_n} dm \right)^{1/p} \\ = \lim_{N \rightarrow \infty} \sum_n^N \left( m(F_n) \right)^{1/p} \\ \leq \lim_{N \rightarrow \infty} \sum_n^N \left( e^{-n} \right)^{1/p} < \infty \end{align*}

Where the last inequality comes from the fact that this is a geometric series with $\frac{1}{e^{\frac{n}{p}}} < 1$ for $n$ large enough so it converges.

Edit

For the second function note that there exists an $N_0$ such that for $n > N_0: \Big | \frac{n^t}{e^n} \Big | < 1$. Then $$\lim_{N \rightarrow \infty} \sum_{n=1}^N \frac{n^t}{e^n} = \sum_{n=1}^{N_0} \frac{n^t}{e^n} + \lim_{N \rightarrow \infty} \sum_{n={N_0}}^N \frac{n^t}{e^n}$$

-
From the line where $\chi \in \{0,1\}$ to the next one your $1/p$ suddenly gets all the sum instead of each indiviual terms. You're getting at the same place as I am in the end. –  Patrick Da Silva Nov 13 '11 at 19:42
@PatrickDaSilva: thanks, I had corrected some other typos earlier, thought I had gotten them all : S –  Rudy the Reindeer Nov 13 '11 at 19:46