Consider a sequence of Lebesgue measurable sets $(M_i)$, such that $M_i\cap M_j=\emptyset \forall i\neq j$. Prove that $\sum_i a_i\chi_{M_i}$ converges in the $L_p$ spaces if and only if $\sum_n |a_i|^pm(M_i)<\infty.$

Not sure where to start on this one. A hint to begin would be helpful. The second part of the statement kind of looks like the norm that is used in $L_p$ space (not exactly but somewhat similar if that is relevant?), i.e, $$||f||_p=\left(\int_E |f|^p\right)^{1/p}$$

But I don't see the relevance of the sets being disjoint. Can someone give me a prod in the right direction?

  • $\begingroup$ The statement of the hypotheses is confused. Do you really mean to say that the intersection of the first $n$ sets is empty, assuming nothing about $M_j$ for $j>n$? You say you don't see the relevance of the sets being disjoint; I don't see why you think they are disjoint... $\endgroup$ – David C. Ullrich Aug 7 '16 at 23:01
  • $\begingroup$ @DavidC.Ullrich: Perhaps I misunderstood something from the original question. Here it is in its exact form: Let $M_i$ be a sequence of disjoint measurable sets... $\endgroup$ – Marshant Aug 7 '16 at 23:20
  • $\begingroup$ That means $M_j\cap M_k=\emptyset$ whenever $j\ne k$. (Which makes the problem easy. Hint: Show the partial sums are a Cauchy sequence.) You should really correct the question. $\endgroup$ – David C. Ullrich Aug 7 '16 at 23:22

Here's a further hint based of David C Ullrich's comment, in case you need it:

Fix $n<m\in\mathbb{Z}^+$. Then:

$$\sum_{i=1}^n \alpha_i\chi_{M_i}-\sum_{i=1}^m\alpha_i\chi_{M_i}=\sum_{i=n+1}^m\alpha_i\chi_{M_i}$$

By disjointness, we see that this function takes the value $\alpha_i$ on $M_i$, and $0$ on the complement of $\bigcup_{i=n+1}^m M_i$.



and hence, by the same token:

$$\left\| \sum_{i=1}^n \alpha_i\chi_{M_i}-\sum_{i=1}^m\alpha_i\chi_{M_i}\right\|_p^p=\int \left|\sum_{i=n+1}^m\alpha_i\chi_{M_i}\right|^p \ dm=\int\sum_{i=n+1}^m |\alpha_i|^p\chi_{M_i}\ dm=\sum_{i=n+1}^m|\alpha_i|^pm(M_i)$$

EDIT: Regarding the usefulness of the disjointness of the $M_i$, call the entire space $X$. If $x\in X$, there's two possibilities: either $x\in \bigcup_{i=n}^mM_i$ or $x\not \in \bigcup_{i=n}^m M_i$. In the latter case, $\sum_{i=n}^m \alpha_i \chi_{M_i}(x)=0$ since each summand is $0$.

Now, if $x\in \bigcup_{i=n}^m M_i$, then there exists $j\in \{n,n+1,\dots,m\}$ such that $x\in M_j$. Furthermore, since the $\{M_i\}$ are disjoint, said number $j$ is unique.

We've thus proved:

$$ \sum_{i=n}^m\alpha_i \chi_{M_i}(x)=\begin{cases} \alpha_j & \text{if $x\in M_j$}\\ 0 & \text{if $x\not\in \bigcup_{i=n}^m M_i$} \end{cases} $$


$$ \left(\sum_{i=n}^m\alpha_i \chi_{M_i}(x)\right)^p=\begin{cases} \alpha_j^p & \text{if $x\in M_j$}\\ 0 & \text{if $x\not\in \bigcup_{i=n}^m M_i$} \end{cases} $$

  • $\begingroup$ Can you elaborate a bit? I don't see where we are using the fact the sets are disjoint? $\endgroup$ – Marshant Aug 9 '16 at 15:54
  • $\begingroup$ @Marshant see my edit. $\endgroup$ – Reveillark Aug 9 '16 at 16:11

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