Probability Theory, Symmetric Difference I'm trying to show this property of the symmetric difference between two sets defined for two sets in a universe $A$ and $B$ by
$$
A\Delta B=(A\cap B^{c})\cup(B\cap A^{c})
$$
I need to show that 
$$
\mathbb{P}(A\Delta C)\leq \mathbb{P}(A\Delta B)+\mathbb{P}(B\Delta C)
$$
for sets $A, B,$ and $C$ in the universe.
I showed in the first part of the problem that 
$$
\mathbb{P}(A\Delta B)=\mathbb{P}(A)+\mathbb{P}(B)-2\mathbb{P}(A\cap B)
$$
My idea was to note that
$$
\mathbb{P}(A\Delta C)\leq\mathbb{P}(A\cap C^{c})+\mathbb{P}(C\cap A^{c})
$$
by probability laws and then leverage the fact that for any set I can write it as a union with another set. That is
$$
\mathbb{P}(A)=\mathbb{P}(A\cap B)+\mathbb{P}(B^{c}\cap A)
$$
and likewise for $C$ to substitute in for $P(A)$ and $P(B)$ terms. However, I end up running in circles. My TA did say I was on the right track, though. Any suggestions would be helpful. Thanks.
 A: Simply use the following formula:
$$
(A\bigtriangleup C) \subset (A\bigtriangleup B) \cup (B\bigtriangleup C)
$$
The proof is as follow:
\begin{align}
A\bigtriangleup C &=(A\cap C^c)\cup(C\cap A^c)
\\
&=((A\cap C^c)\cup(C\cap A^c))\cap (B\cup B^c)
\\
&=(A\cap C^c\cap B)\cup(C\cap A^c\cap B)\cup (A\cap C^c\cap B^c)\cup(C\cap A^c\cap B^c)
\\
&\subset (C^c\cap B)\cup(A^c\cap B)\cup (A\cap B^c)\cup(C\cap B^c)
\\
&=((A\cap B^c)\cup(A^c\cap B))\cup((C^c\cap B)\cup(C\cap B^c))
\\
&=(A\bigtriangleup B) \cup (B\bigtriangleup C)
\end{align}
A: If $\chi_U$ denotes the characteristic function of the set $U$, then we have $\chi_{A\Delta B} = |\chi_A - \chi_B|$. Hence we have 
\begin{align*}
\mathbb{P}(A\Delta C) &= \int |\chi_A - \chi_C| d\mu \\
&= \int |(\chi_A - \chi_B) + (\chi_B - \chi_C)| d\mu \\ 
&\leq \int (|\chi_A - \chi_B| + |\chi_B - \chi_C|) d\mu \\
&= \int (|\chi_A - \chi_B|d\mu + \int |\chi_B - \chi_C|) d\mu\\
&= \mathbb{P}(A\Delta B) +  \mathbb{P}(B\Delta C) 
\end{align*}
A: By what you proved, we can write the inequality that we want to show as 
$$
P(A) + P(C) - 2 P(A \cap C) \leq P(A) + P(B) - 2 P(A \cap B) + P(B) + P(C) - 2 P(B \cap C).
$$
Subtracting redundant terms, we see that this inequality is equivalent to 
$$
- 2 P(A \cap C) \leq 2 P(B) - 2 P(A \cap B) - 2 P (B \cap C).
$$
Then, dividing by 2, 
$$
-P(A \cap C) \leq P(B) - P(A \cap B) - P(B \cap C).
$$
So, it suffices to show 
$$
P(B) + P(A \cap C) - P(A \cap B) - P(B \cap C) \geq 0.
$$
Now, using your trick on $P(B)$, we can write this inequality as 
$$
P(A \cap B) + P(A^c \cap B) + P(A \cap C) - P (A \cap B) - P(B \cap C) \geq 0 
$$
and then cancelling, 
$$
P(A^c \cap B) + P(A \cap C) - P(B \cap C) \geq 0.
$$
This inequality is true because 
$$
B \cap C = \{A^c \cap [B \cap C] \} \cup \{A \cap [B \cap C]\} \subseteq \{A^c \cap B\} \cup \{A \cap C\}.
$$
Indeed, because of the above, we have by the monotonicity of probability, 
$$
P(B \cap C) \leq P(\{A^c \cap B\} \cup \{A \cap C\}) \leq P(A^c \cap B) + P(A \cap C).
$$
