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Let $(\Omega, \mathcal{F}, \textbf{P})$ be a probability space and $A,B, A_i$ events in $\mathcal{F}$. Prove the following properties of every probability measure:

  • Monotonicity: If $A \subseteq B$ then $\textbf{P}(A) \leq \textbf{P}(B)$.
  • Sub-additivity: If $A \subseteq \bigcup_{i} A_i$ then $\textbf{P}(A) \leq \sum_{i} \textbf{P}(A_i)$.
  • Continuity from Below: If $A_{i} \uparrow A$, then $\textbf{P}(A_i) \uparrow \textbf{P}(A)$.
  • Continuity from Above: If $A_{i} \downarrow A$, then $\textbf{P}(A_i) \downarrow \textbf{P}(A)$.

For the first, one could consider $\textbf{P}(B-A)$? For the second, one could consider $\textbf{P}(A_{i}-A)$ over all $i$? It seems like the third and fourth properties follow from the first two. In particular, if $A_{1} \subseteq A_2 \subseteq \dots$ and $\bigcup_{i} A_i = A$, then $\textbf{P}(A_1) \leq \textbf{P}(A_2) \leq \dots$ and $\sum_{i} \textbf{P}(A_i) = \textbf{P}(A)$.

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Am I mistaken, or is this a standard stuff found in each measure theory textbook? – mpiktas Dec 31 '10 at 19:36
@mpiktas: This is a standard question. It is typically the first exercise question once the definition of measure is introduced. – user17762 Dec 31 '10 at 19:41
can you please specify what kind of answers are you expecting, since these are standard measure theory questions. – mpiktas Dec 31 '10 at 20:42
up vote 5 down vote accepted

Monotonicity follows from the countable additivity of measure. You need to recognize whenever $A \subseteq B$, we have $B = A \cup (B \cap A^{c})$. And note that $A$ and $B \cap A^{c}$ are disjoint sets.

Sub-additivity is proved as follows. First prove that $\mathbf{P}(\displaystyle \cup_{i=1}^{\infty} A_i) \leq \sum_{i=1}^{\infty} \mathbf{P}(A_i)$.

This can be proved by re-writing $\displaystyle \cup_{i=1}^{\infty} A_i$ as a disjoint union and then use the countable additivity of a measure to get the desired result. Now use monotonicity proved in the previous part to conclude the desired result.

What, continuity from below means is "If we have a nested sequence of sets, i.e. $A_1 \subseteq A_2 \subseteq \ldots $ and $\cup_{i=1}^{\infty} A_i = A$, then we have continuity from below i.e. $\displaystyle \lim_{n \rightarrow \infty} \mathbf{P}(A_n) = \mathbf{P}(A)$".

The trick here again is to rewrite $\displaystyle \cup_{i=1}^{\infty} A_i$ as a disjoint union and massage around to get the desired result.

Similarly, continuity from above means, "If we have a sequence of sets such that $A_{1} \supseteq A_{2} \supseteq \ldots $ and $\displaystyle \cap_{i=1}^{\infty} A_i = A$, then we have continuity from above i.e. $\displaystyle \lim_{n \rightarrow \infty} \mathbf{P}(A_n) = \mathbf{P}(A)$".

This follows from previous part when we consider the sequence of sets $B_i = A_1 \backslash A_i$ and noting that $\mathbb{P}(A_1) \leq 1 < \infty$.

Note: In general, all these properties except the last one hold true for any measure. The last property holds true if at-least one of the sets in the sequence we consider has a finite measure.

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