# Using Lebesgue dominated convergence theorem

Let $A_1,A_2,A_3...$ be measurable sets. Let $m \in N$ and let $E_m$ be the set defined as follows: $x \in E_m$ iff x is a member of at least m of the sets $A_k$ Prove that $E_m$ is a measurable set and that ,

$m.\lambda(E_m) \leq \sum_{k=1}^{infty} \lambda(A_k)$

MY approach : To prove that the following set is measurable I was hoping to use the caratheodary characterisation of the measurable sets , however I failed to do so. Maybe there is a simpler way to prove that the above set is measurable.

I was hoping to use the lebesgue dominated convergence theorem to prove the above inequality on the function For that I identified that $E_m$'s form a decreasing sequence of sets , and if I define $f_m$ to be the characteristic functions of the sets $E_m$ maybe I could get further , as then I would be able to use the $f_1$ to be the function which bounds all the f's to apply the dominated convergence theorem . Any input on validity of the ideas will be appreciated .

Since I am self studying this , it would be great if you could show me how to translate my ideas in a formal mathematical proof. So please be kind enough to atlas provide the basic structure of the proof.

To see that $E_m$ is measurable let $\Lambda_m=\{ L\subset\Bbb N\mid |L|=m\}$ be the subsets of $\Bbb N$ that have $m$ elements. Note that this is a countable set. Then $$E_m=\bigcup_{L\in \Lambda_m}\left(\bigcap_{k\in L}A_k\right)$$ is a countable union of measurable sets and thus measurable.

The Lebesgue dominated convergence theorem won't necessarily help you and I can't see how you can apply it in a way that is not artificial. Rather you can see that $E_m\subseteq E_1 = \bigcup_k A_k$ and for that reason $$\lambda(E_m)\leq\lambda(E_1)≤\sum_k\lambda(A_k)$$ by monotonicity and sub-addititivity of the measure.

• The inequality that you obtain does not have a m on the RHS . – Noob101 Mar 23 '17 at 20:04
• Could you help me see how $Em=⋃L∈Λm(⋂k∈LAk$ – Noob101 Mar 23 '17 at 20:11
• Suppose a point $p$ lies in $m$ different $A_k$. This means there exists $A_{k_1},...,A_{k_m}$ so that $p\in \bigcap_{i=1}^m A_{k_i}$. It follows $p$ lies in the right hand side. On the other hand every point in the right hand side lies in the intersection of $m$ different $A_k$, thus also in $E_m$. – s.harp Mar 24 '17 at 9:18
• Now you always have $E_m\subset E_1$, since $E_1=\bigcup_k A_k$. Measures are monotone, this means that if $X\subset Y$ that then $\lambda(X)≤\lambda(Y)$. This implies the first inequality. The second follows from $\sigma$ sub-additivity, for any countable union you have $\lambda(\bigcup_n X_n)≤\sum_n\lambda(X_n)$. – s.harp Mar 24 '17 at 9:19