# Measure theory problem.

Suppose $X, M, \mu$ is a measure space and $A_1, A_2, A_3, ...$ are sets in $M$ such that each point of $X$ belongs to no more than d of the $A_k$ 's. then

$\sum^{\infty}_{k=1} \mu(A_k) \le d \mu (\bigcup^{\infty}_{k=1} A_k)$

Plz help to prove it. It's hard to me

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What have you considered so far? –  mixedmath May 6 '12 at 19:23
Someone's got a homework assignment due tomorrow? –  Eric Gregor May 6 '12 at 19:25
OP has posted 9 questions today with no work and some variation of "Plz help, this is hard". –  Eric Gregor May 6 '12 at 19:28
Yes, and some even have the same title. Please, show a little effort. –  mixedmath May 6 '12 at 19:35
In the long run, you need to learn to do your homework yourself. It is fine to ask for help, we all do at time, but try it yourself first, share what you tried, and when you ask, listen to the replies... –  Chris K. Caldwell May 6 '12 at 21:44

If $A_k$ are pairwise disjoint, i.e. $A_k \cap A_i = \varnothing$ if $i \neq k$, then you have $$\sum_k \mu (A_k) = \mu (\bigcup_k A_k)$$
Next consider the case where each point $x$ in $X$ is in exactly $d$ of the $A_k$. In this case you count the measure of each $A_k$ $d$-times too many in the sum, hence $$\sum_k \mu (A_k) = d \mu (\bigcup_k A_k)$$
Now you see that if each point is in at most $d$ of the $A_k$, you count the measure of each $A_k$ at most $d$-times too many, hence $$\sum_k \mu (A_k) \leq d \mu (\bigcup_k A_k)$$