How to show that $P_1 \wedge P_2 \neq 0$? Suppose that $P_1$ and $P_2$ are finitely additive probability measures on $(X, \mathcal{B})$, and suppose that there's some $ B_0 \in \mathcal{B}$ such that $0 < P_1(B_0) < 1$ and $P_2(B_0) = 1$.
I'd like to show that $P_1 \wedge P_2 \neq 0$. I know that $P_1 \wedge P_2$ is defined for all $B \in \mathcal{B}$ by 
$$(P_1 \wedge P_2)(B) = \inf\{P_1(A) + P_2(B-A): A \subseteq B, A \in \mathcal{B} \},$$
and from this it is clear that $P_1 \wedge P_2 \geq 0$, since $P_1$ and $P_2$ are probabilities.
To finish, I just need to show that there's some $B \in \mathcal{B}$ such that $(P_1 \wedge P_2)(B) > 0$. I have tried to show this for $B_0$ but am unable to conclude.
As a follow-up, if $P_1$ and $P_2$ are mutually singular probabilities, then is $P_1 \wedge P_2 = 0$?
 A: Recall that probabilities are mutually singular, if there exists $B_0\in\mathcal B$ such that $P_1(B_0)=0$ and $P_2(X\setminus B_0)=0$.
So, the probabilities are not mutually singular, if for any $B_0\in\mathcal B$, if $P_1(B_0)=0$, then $P_2(X\setminus B_0)>0$, or equivalently $P_2(B_0)<1$.
Take next $B=X$. For any $A\subseteq X$, $A \in \mathcal{B}$, either $P_1(A)>0$ or $P_1(A)=0$, $P_2(X\setminus A)>0$.  
Then
$$(P_1 \wedge P_2)(X) = \inf\bigl\{P_1(A) + P_2(X\setminus A): A \subseteq X, A \in \mathcal{B} \bigr\}>0.$$
As user435571 noted in comments below, we need to consider the sad situation when this $\inf$ is newertheless equal to zero. If it is so, then there exists a sequence of sets $A_n\subseteq X$, $A_n\in\mathcal B$ s.t. 
$$P_1(A_n)\to 0,\ P_2\left(\overline A_n\right)=P_2(X\setminus A_n)\to 0.$$
Without loss of generality, we can assume that $P_1(A_n)\leq 2^{-n}$ for any $n$. Really, we can assign the number $1$ to the first event in this sequence which has the probability $P_1$ less than $1/2$. Then assign number $2$ to the first event among the others whose probability is less than $1/4$ and so on.
Next define new sequence of events $B_n\in\mathcal B$ as follows:
$$
B_n=\bigcup_{k=n}^\infty A_k, \quad \overline B_n=\bigcap_{k=n}^\infty \overline A_k.
$$
$B_n$ form a decreasing family: 
$$B_1\supseteq B_2\supseteq B_3\supseteq \ldots,$$
and the family of $\overline B_n$ is increasing:
$$\overline B_1\subseteq \overline B_2\subseteq \overline B_3\subseteq \ldots.$$
Using continuity property of probability measures, one get:
$$\tag{1}\label{1}
P_1(B_n) \to P_1\left(\bigcap_{n=1}^\infty B_n\right) = P_1\left(\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k\right) = P_1(\{A_n \text{ i.o}\}).$$
Here the event $\mathbb A=\{A_n \text{ i.o}\}$ means that in the sequence $\{A_n\}$ the events appears infinitely often.
But 
$$\tag{2}\label{2}
P_1(B_n) = P_1\left(\bigcup_{k=n}^\infty A_k\right)\leq \sum_{k=n}^\infty P_1(A_k)\leq \sum_{k=n}^{\infty} 2^{-k}=2^{-n+1}\to 0.
$$
Conclude from (\ref{1}) and (\ref{2}) that $P_1(\mathbb A)=0$.
Next, the continuity of probability measure implies that
$$
P_2(\overline B_n)\to P_2\left(\bigcup_{n=1}^\infty \overline B_n\right)=P_2\left( \overline{\bigcap_{n=1}^\infty B_n}\right)=P_2\left(\overline{\mathbb A}\right).
$$
From the other side, $P_2(\overline B_n)\leq P_2(\overline A_n)\to 0$. Conclude that $P_2\left(\overline{\mathbb A}\right)=0$.
We have found the event $\mathbb A=\{A_n \text{ i.o}\}$ such that $P_1(\mathbb A)=0$ and $P_2(X\setminus {\mathbb A})=0$, which contradicts the assumption that our measures are not mutually singular.
This contradiction shows that $(P_1\wedge P_2)(X)>0$.
And the converse follows simply too. Let there exists $B_0\in\mathcal B$ such that $P_1(B_0)=0$ and $P_2(X\setminus B_0)=0$. Then for any $B\in\mathcal B$,
$P_1(B\cap B_0)=0$, $P_2(B\setminus (B\cap B_0))=0$ and then $(P_1 \wedge P_2)(B)=0$.
