Measure Union of n Measurable Sets: Formula & Examples

[mathmari]

Hey!

At any metric space, find a formula that gives the measure of the union of $n$ measurable sets, not necessary disjoint.

If the sets are disjoint the measure of the union is $$\mu \left ( \cup_{n=1}^{\infty} A_n \right)=\sum_{n=1}^{\infty}\mu(A_n)$$ right??

And when the sets are not disjoint the $=$ gets $\leq$.

Is this the formula that I am asked to find?? (Wondering)

 

[Evgeny.Makarov]

At any metric space, find a formula that gives the measure of the union of $n$ measurable sets, not necessary disjoint.

One option is
\[
\mu\left(\bigcup_{i=1}^n A_i\right)=\sum_{i=1}^n\mu\left(A_i\setminus\bigcup_{k=1}^{i-1}A_k\right).
\]

 

[Euge]

Since you are asked to find a formula, you must give an equation, not an inequality. I'll write down the formula, and leave it to you to prove it.

Let $A_1,\ldots, A_n$ be $n$ measurable sets. Then

$$ \mu(\bigcup_{j = 1}^n A_j) = \sum_{j = 1}^k \mu(A_j) -\sum_{1\le j_1 < j_2 \le n} \mu(A_{j_1} \cap A_{j_2}) +\sum_{1 \le j_1 < j_2 < j_3 \le n}\mu(A_{j_1} \cap A_{j_2} \cap A_{j_3}) - \cdots + (-1)^{n-1} \mu(\bigcap_{j = 1}^n A_j)$$

 

[mathmari]

One option is
\[
\mu\left(\bigcup_{i=1}^n A_i\right)=\sum_{i=1}^n\mu\left(A_i\setminus\bigcup_{k=1}^{i-1}A_k\right).
\]

Since you are asked to find a formula, you must give an equation, not an inequality. I'll write down the formula, and leave it to you to prove it.

Let $A_1,\ldots, A_n$ be $n$ measurable sets. Then

$$ \mu(\bigcup_{j = 1}^n A_j) = \sum_{j = 1}^k \mu(A_j) -\sum_{1\le j_1 < j_2 \le n} \mu(A_{j_1} \cap A_{j_2}) +\sum_{1 \le j_1 < j_2 < j_3 \le n}\mu(A_{j_1} \cap A_{j_2} \cap A_{j_3}) - \cdots + (-1)^{n-1} \mu(\bigcap_{j = 1}^n A_j)$$

I understand! Thank you both very much! (Sun)