Show that $\mu(A)=\sup \left\{\mu_1(B)+\mu_2(A \setminus B) : B \subset A \right\}$ is measure.

Question

Let $(X ,\mathfrak{m})$-measurable space. Let $\mu_1$ and $\mu_2$ will be measures. For $A \in \mathfrak{m}$ define : $\mu(A)=\sup \left\{\mu_1(B)+\mu_2(A \setminus B) : B \subset A \right\} $ Show that $\mu$ is measure.

1.$\mu(\emptyset)=0$

This is obvious.

2.$\mu(\bigcup_{n=1}^{\infty} A_n)=\sum_{n=1}^{\infty}\mu(A_n)$

If $B=\emptyset$ (obvious). What, if $B \neq \emptyset$ ?

Answer 1

Fix a pairwise disjoint sequence $\{A_n\}$.

Let $B\subset \bigcup_nA_n$. Then \begin{align} \mu_1(B)+\mu_2\left(\bigcup_nA_n\setminus B\right) &=\mu_1\left(\bigcup_n(A_n\cap B)\right)+\mu_2\left(\bigcup_n(A_n\setminus B)\right)\\ \ \\ &=\sum_n\mu_1(A_n\cap B)+\sum_n\mu_2(A_n\setminus B)\\ \ \\ &=\sum_n (\mu_1(A_n\cap B)+\mu_2(A_n\setminus B))\\ \ \\ &=\sum_n (\mu_1(A_n\cap B)+\mu_2(A_n\setminus (A_n\cap B)))\\ \ \\ &\leq\sum_n\mu(A_n). \end{align} As $B$ was arbitrary, we obtain $$\tag1 \mu\left(\bigcup_nA_n\right)\leq\sum_m\mu(A_n). $$

For the reverse inequality, fix $\varepsilon>0$, and for each $n$ choose $B_n\subset A_n$ such that $$ \mu(A_n)\leq \mu_1(B_n)+\mu_2(A_n\setminus B_n)+\frac\varepsilon{2^n}. $$ Let $B=\bigcup_n B_n$. Note that all the $B_n$ are pairwise disjoint, since the $A_n$ are pairwise disjoint. Then \begin{align} \sum_n\mu(A_n) &\leq \sum_n \left(\mu_1(B_n)+\mu_2(A_n\setminus B_n)+\frac\varepsilon{2^n} \right)\\ \ \\ &=\varepsilon+\sum_n (\mu_1(B_n)+\mu_2(A_n\setminus B_n))\\ \ \\ &=\varepsilon + \mu_1(B)+\sum_n (\mu_2(A_n\setminus B))\\ \ \\ &=\varepsilon + \mu_1(B)+\mu_2\left(\bigcup_n (A_n\setminus B)\right)\\ \ \\ &=\varepsilon + \mu_1(B)+\mu_2\left(\bigcup_n A_n\setminus B\right)\\ \ \\ &\leq\varepsilon + \mu\left(\bigcup_n A_n\right). \end{align} As we can do this for all $\varepsilon>0$, the inequality holds without $\varepsilon$ and so $$\tag2 \sum_m\mu(A_n)\leq\mu\left(\bigcup_nA_n\right) $$

Answer 2

By the definition of $\operatorname{sup}$, for a sequence $\{\epsilon_n\}$, where $\epsilon_n > 0$, we may find a $B_n$ corresponding to $\epsilon_n$ and $A_n$ so that $$\mu(A_n) < \mu_1(B_n) + \mu_2(A_n\setminus B_n) + \epsilon_n < \mu(A_n) + \epsilon_n$$ Note that, given that $\{A_n\}$ are disjoint, hence so are $\{B_n\}$ and $\{A_n\setminus B_n\}$, so apply the fact that $\mu_1,\mu_2$ are measures:

\begin{align} \sum_{n=1}^{\infty} \mu(A_n) &\leq \sum_{n=1}^{\infty}\left(\mu_1(B_n) + \mu_2(A_n\setminus B_n) + \epsilon_n\right)\\ &= \mu_1\left(\bigcup_{n=1}^{\infty} B_n\right) + \mu_2\left(\bigcup_{n=1}^{\infty} (A_n\setminus B_n)\right) + \sum_{n=0}^{\infty}\epsilon_n \\ &= \mu_1\left(\bigcup_{n=1}^{\infty} B_n\right) + \mu_2\left(\bigcup_{n=1}^{\infty} A_n\setminus \bigcup_{n=1}^{\infty} B_n\right) + \sum_{n=0}^{\infty}\epsilon_n\\ &\leq \sup\left\{\,\mu_1(B) + \mu_2\left(\bigcup_{n=1}^{\infty}A_n \setminus B\right) \colon B \subset \bigcup_{n=1}^{\infty}A_n\,\right\} + \sum_{n=0}^{\infty}\epsilon_n\\ &= \mu\left(\bigcup_{n=0}^{\infty}A_n\right) + \sum_{n=0}^{\infty}\epsilon_n\\ \end{align}

Now we can make the summation on the right as small as desired: Take $\epsilon > 0$ then if we initially choose $\{\epsilon_n\}$ so that $\sum_{n=0}^{\infty} \epsilon_n \leq \epsilon$, e.g., by taking $\epsilon_n=\frac{\epsilon}{2^{n}}$, then $$\sum_{n=1}^{\infty} \mu(A_n) \leq \mu\left(\bigcup_{n=0}^{\infty}A_n\right) + \epsilon $$ for any $\epsilon > 0$, hence $$\sum_{n=1}^{\infty} \mu(A_n) \leq \mu\left(\bigcup_{n=0}^{\infty}A_n\right)$$

The other way around: take $A_1,A_2$ disjoint and fix $\epsilon > 0$, then there exists some $B \subset A_1\cup A_2$ so that

\begin{align} \mu(A_1\cup A_2) &< \mu_1(B) + \mu_2((A_1\cup A_2) \setminus B) + \epsilon\\ &= \mu_1(B\cap A_1) + \mu_1(B\cap A_2) + \mu_2(A_1\setminus \underbrace{B}_{= B\cap A_1}) + \mu_2(A_2\setminus \underbrace{B}_{= B\cap A_2} ) + \epsilon\\ &\leq \mu(A_1) + \mu(A_2) + \epsilon \end{align} as this holds for all $\epsilon$, $\mu(A_1\cup A_2) \leq \mu(A_1) + \mu(A_2)$ which inductively implies $$\mu\left(\bigcup_{n=0}^{\infty}A_n\right) \leq\sum_{n=1}^{\infty} \mu(A_n)$$

(Really we could have just used the fact that $\operatorname{sup}$ is sub-additive, but you would prove that using an $\epsilon$-argument as above anyway).