Example of measure space $(X,\mathcal B,\mu)$ and measurables and disjoint sets $A,B\in \mathcal B$ , such that: $\mu(A) = \mu(B) = \mu(X)$

Question

I need to give example of measure space $(X,\mathcal B,\mu)$ and measurable and disjoint sets $A,B\in \mathcal B$ , such that: $\mu(A) = \mu(B) = \mu(X)$

I saw the following solution:

Let $A=[0,\frac12]$ and $B=[\frac12,1]$, then $A$ and $B$ are disjoint, measurable sets in $\mathcal{B}$ and $\mu(A)=\mu(B)=1/2=\mu(X)$.

But I don't really understand why this is true, can someone explain the solution to me?

Answer 1

If $A$ and $B$ are disjoint and measurable, then by definition of a measure $\mu$ we have $$\mu(A\cup B)=\mu(A)+\mu(B).$$ Therefore if we want to have $A,B\subseteq X$ such that $$\mu(A)=\mu(B)=\mu(X)$$ since $A\cup B\subseteq X$ and therefore $\mu(A\cup B)\leq \mu(X)$ we would either have $$\mu(A)=\mu(B)=\mu(X)=\infty$$ or $$\mu(A)=\mu(B)=\mu(X)=0.$$ The example you gave fulfills neither of those conditions. Try to think of measurable disjoint sets $A,B$ in $\mathbb R$ with infinite Lebesgue measure.

Answer 2

Let $C$ be the Cantor set, which has measure $0$ and $D$ be $C$ shifted up by $1$, i.e. $$ D=\{1+x\ |\ x\in C\}. $$

Take $X=C\cup D$, then $\mu(C)=\mu(D)=\mu(X)=0$.