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

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

 

[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

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

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

 

[blue_raver22]

Hi there! Yes, you are correct. The formula for the measure of the union of $n$ measurable sets, whether they are disjoint or not, is given by:

$$\mu \left(\bigcup_{i=1}^{n} A_i \right) = \sum_{i=1}^{n}\mu(A_i)$$

This formula holds true for any metric space, as long as the sets are measurable. When the sets are disjoint, the measure of the union is simply the sum of the measures of each set. However, when the sets are not disjoint, the measure of the union may be less than or equal to the sum of the measures. This is because there may be some overlap between the sets, and this overlap is counted multiple times when adding the measures separately.

Let me give you an example to better illustrate this. Consider three sets $A$, $B$, and $C$ in a metric space, where $A$ and $B$ are disjoint but $C$ overlaps with both $A$ and $B$. If we add the measures of each set, we get $\mu(A) + \mu(B) + \mu(C)$. However, this will overestimate the measure of the union, as the overlap between $A$ and $C$ and between $B$ and $C$ is counted twice. So, the actual measure of the union is $\mu(A) + \mu(B) + \mu(C) - \mu(A \cap C) - \mu(B \cap C)$.

I hope this helps clarify the formula for the measure of the union of $n$ measurable sets. Let me know if you have any other questions!