Exercise 1.4.20 (Properties of finitely additive measure)

Question

Let $\mu:\mathcal{B}	o[0,+\infty]$ be a finitely additive measure on a Boolean algebra $\mathcal{B}$. Establish the following facts:
\begin{enumerate}[label=(oman*)]
\item (Monotonicity) If $E,F$ are $\mathcal{B}$-measurable and $E\subseteq F$ then $\mu(E) \leq \mu(F)$.
\item (Finite additivity) If $E_1,\ldots,E_k$ are $\mathcal{B}$-measurable and disjoint, then $\mu(\bigcup_{n=1}^{k}E_n) = \sum_{n=1}^{k}\mu(E_n)$.
\item (Finte subadditivity) If $E_1,\ldots,E_k$ are $\mathcal{B}$-measurable, then $\mu(\bigcup_{n=1}^{k}E_n) \leq \sum_{n=1}^{k}\mu(E_n)$.
\item (Finite inclusion-exclusion principle) If $E,F$ are $\mathcal{B}$-measurable, then $\mu(E \cup F) = \mu(E) + \mu(F) - \mu(E\cap F)$.
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}[label=(oman*)]
\item 	extit{(Monotonicity)} Notice that the sets $F/E$ and $F \cap E = E$ are disjoint, and $(F/E) \cup E = F$. By finite additivity we then have $\mu(F) = \mu(F/E) + \mu(E) \geq \mu(E)$, where we have used the non-negativity of $\mu:\mathcal{B}	o[0,+\infty]$ in the last step.
  
\item 	extit{(Finite additivity)} The induction base case $n=2$ is true by definition, and in the induction step we have $$\mu\left(\bigcup_{n=1}^{k+1}E_night) = \mu\left(\bigcup_{n=1}^{k}E_n \cup E_{k+1}ight) =\mu\left(\bigcup_{n=1}^{k+1}E_night)+\mu(E_{k+1})= \sum_{n=1}^{k}\mu(E_n) + \mu(E_{k+1}) =  \sum_{n=1}^{k+1}\mu(E_n)$$
  
\item 	extit{(Finte subadditivity)} The induction base for $n=2$ follows by
  $$\mu(E\cup F) = \mu\left((E/F)\cup(F/E)\cup(E\cap F)ight) = \mu((E/F)\cup(E \cap F)) + \mu(F/E) \leq \mu(E) + \mu(F)$$
  from the finite additivity and monotonicity properties. The induction step follows in a similar spirit:
  \begin{align*}
    \mu\left(\bigcup_{n=1}^{k+1}E_night) &= \mu\left(\bigcup_{n=1}^{k}E_n \cup E_{k+1}ight) = \mu\left(\left[\bigcup_{n=1}^{k+1}E_n/E_{k+1}ight] \cup \left[E_{k+1}/\bigcup_{n=1}^{k+1}E_night] \cup \left[\bigcup_{n=1}^{k+1}E_n\cap E_{k+1} ight]ight)\
    &= \mu\left(\left[\bigcup_{n=1}^{k+1}E_n/E_{k+1}ight] \cup \left[\bigcup_{n=1}^{k+1}E_n\cap E_{k+1} ight]ight)+\mu\left( E_{k+1}/\bigcup_{n=1}^{k+1}E_night)\
    &\leq \mu\left(\bigcup_{n=1}^{k+1}E_night)+\mu\left( E_{k+1}ight)\
    &\leq \sum_{n=1}^{k}\mu(E_n/E_{k+1}) + \mu(E_{k+1}/\bigcup_{n=1}^{k+1}E_n) \leq  \sum_{n=1}^{k+1}\mu(E_n)
  \end{align*}
	
\item 	extit{(Finite inclusion-exclusion principle)} We have
  \begin{align*}
    \mu(E) &= \mu(E/F) + \mu(E\cap F)\
    \mu(F) &= \mu(F/E) + \mu(F \cap E)
  \end{align*}
  The last term being the same in both equations. Adding these equations up we obtain
  \begin{align*}
    \mu(E) + \mu(F) = \mu(E/F) + \mu(F/E) + \mu(E \cap F) + \mu(E \cap F) = \mu(E \cup F) + \mu(E\cap F)
  \end{align*}
  since we have the union $E \cup F = (E/F) \cup (F/E) \cup (E \cap F)$, and this representation is disjoint.
\end{enumerate}

Exercise 1.4.20 (Properties of finitely additive measure)

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Exercise 1.4.20 (Properties of finitely additive measure)

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