# Generalization of subadditivity for a probability function

This is an exercise from Resnick's "A probability path" book (chapter 2, exercise 3).

We have a $$(\Omega, \mathcal{B}, P)$$ probability space with $$B_i \subset A_i$$. The generalization of subadditivy (and what I need to prove) is (the unions and sums are all of countable collection of sets):

$$P(\cup A_i) - P(\cup B_i) \leq \sum_i (P(A_i) - P(B_i))$$

I tried using the monotonicity property and work with the following inequalities

• $$P(B_i) \leq P(A_i)$$
• $$P(\cup A_i) \leq \sum P(A_i)$$
• $$P(\cup B_i) \leq \sum P(B_i)$$
• $$\sum P( B_i) \leq \sum P(A_i)$$

But couldn't figure out how to prove the statement from this. Then I tried using the same trick for proving subadditivity and constructing disjoint sets, so you can write (using $$+$$ to represent the union):

$$\cup A_i = A_1 + A_2 \cap A_1^c + A_3 \cap A_2^c \cap A_1^c + ...$$

so that

$$P(\cup A_i) \leq P(A_1) + P(A_2) + ...$$

The same can be done with $$B_i$$, so I can write

$$P(\cup A_i) - P(\cup B_i) = P(A_1) + P(A_2 \cap A_1^c) + P(A_3 \cap A_2^c \cap A_1^c) - P(B_1) - P(B_2 \cap B_1^c) - P(B_3 \cap B_2^c \cap B_1^c)$$

and this needs to be compared less or equal than $$P(A_1) - P(B_1) + P(A_2) - P(B_2) + ...$$

I think it would be enough to show (and then do induction for the extra intersections) that

$$P(A_2 \cap A_1^c) - P(B_2 \cap B_1^c) \leq P(A_2) - P(B_2)$$

but I haven't been able to show that. $$P(A_2) \geq P(A_2 \cap A_1^c)$$ but the same is true for the $$B$$ term, so the inequality is not clear.

There might be an easier way to prove this though! Thanks for your time.

• Define $C_i=A_i\setminus B_i$ and re-write the right hand side in terms of the $C_i.$ – whuber Aug 28 '20 at 18:11

• If $$B \subseteq A$$, then $$P(A \setminus B ) = P(A) - P(B)$$.
• Set difference distributes over unions: $$(R \cup S) \setminus T = (R \setminus T) \cup (S \setminus T)$$.
Since $$B_i \subseteq A_i$$ for all $$i$$, it follows that $$\bigcup_i B_i \subseteq \bigcup_i A_i$$, whence \begin{aligned} P\left(\bigcup_i A_i\right) - P\left(\bigcup_i B_i\right) &= P\left(\left(\bigcup_i A_i\right) \setminus \left(\bigcup_i B_i\right)\right) \\ &= P\left(\bigcup_i \left(A_i \setminus \left(\bigcup_j B_j\right)\right)\right). \end{aligned} Now observe that $$A_i \setminus \left(\bigcup_j B_j\right) \subseteq A_i \setminus B_i$$, so we continue: \begin{aligned} P\left(\bigcup_i A_i\right) - P\left(\bigcup_i B_i\right) &= P\left(\bigcup_i \left(A_i \setminus \left(\bigcup_j B_j\right)\right)\right) \\ &\leq P\left(\bigcup_i A_i \setminus B_i\right) \\ &\leq \sum_i P\left(A_i \setminus B_i\right) \\ &= \sum_i \left(P(A_i) - P(B_i)\right). \end{aligned}