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.

  • 1
    $\begingroup$ Define $C_i=A_i\setminus B_i$ and re-write the right hand side in terms of the $C_i.$ $\endgroup$
    – whuber
    Aug 28, 2020 at 18:11

1 Answer 1


There is actually an easier proof using just monotonicity and subadditivity.

Two things to remember before the proof:

  • 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: $$ \begin{aligned} P\left(\bigcup_i A_i\right) - P\left(\bigcup_i B_i\right) &= P\left(\bigcup_i \left(A_i \setminus \left(\bigcup_j B_j\right)\right)\right) \\ &\leq P\left(\bigcup_i A_i \setminus B_i\right) \\ &\leq \sum_i P\left(A_i \setminus B_i\right) \\ &= \sum_i \left(P(A_i) - P(B_i)\right). \end{aligned} $$


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