Intuition behind a probability measure These are some related questions to the question posed here. I don't think it's a duplicate.
What is a probability measure? How do they differ from measure spaces? And could anyone elaborate on the "physical" interpretation of a probability measure? For any measurable subset here, what is its interpretation of its measure?
Can anybody answer these questions by providing some enlightening examples along the way?
 A: A probability measure is a measure space, $(X, \mathcal{M}, \mu)$, such that the entire space $X$ is a measurable set, and $\mu(X) = 1$. Lebesgue measure on $[0, 1]$, for instance, is a probability measure. A second examples is the following. Fix a sequence, $a_1, a_2, \ldots$, of nonnegative numbers whose sum is one. Now let $X$ be the set of positive integers, let $\mathcal{M}$ be all subsets of $X$, and let, for $A \in \mathcal{M}$,$$\mu(A) = \sum_{i \in A} a_i.$$The result is a probability measure (since this is a measure space, and it has $X$ itself measurable, with $\mu(X) = 1$). The similar construction, with $X$ finite, also yields a probability measure.
Since every probability measure is also a measure space, all properties of measure spaces hold also for probability measures. Thus, for example, in a probability measure:


*

*The empty set is measurable.

*Differences, finite unions, and countable intersections of measurable sets are measurable.

*If the union of a countable collection of disjoint measurable sets is measurable, then the measure of that union is the sum of the measures of the sets, etc.


But, by virtue of the extra conditions on a measure space in order that it be a probability measure, there are some further properties special to probability measures. Indeed, let $(X, \mathcal{M}, \mu)$ be a probability measure. Then:


*

*For each measurable set $A$, $A^\text{c}$ (the complement of $A$) is also measurable (since $A^\text{c}$ is the difference of the two measurable sets $X$ and $A$).

*For each measurable set $A$, $\mu(A) \le 1$ (since $A \subset X$, while $\mu(X) = 1$).

*For $A_1, A_2, \ldots$ any countable collection of measurable sets, their union, $\cup A_i$, is also measurable (since these are all subsets of the measurable set $X$).


The physical interpretation of a probability measure is the following. We imagine that we have some experiment, that can be repeated as often as we wish. Then $X$ represents the set of possible outcomes of that experiment. That is, each time the experiment is performed, there will be returned some point of the set $X$. An example is the experiment of drawing a card from a standard deck. In this case, the set $X$ would have $52$ elements, corresponding to the $52$ possible results of this experiment. A measurable subset $A$ of $X$ represents the specification of some particular collection of possible outcomes. In our example, the specification "the card is a diamond" would correspond to the subset $A$ of $X$ consisting of just the diamonds. This is a subset with $13$ elements. The empty subset of $X$ is the specification "no outcome"; and the entire set $X$ the specification "any outcome". For $A$ any measurable subset of $X$, the measurable subset $A^\text{c}$ represents the specification "not in $A$", or "the specification $A$ does not occur". In our example, if $A$ represents the cards that are diamonds, then $A^\text{c}$ represents "the cards that are not diamonds". That is, $A^\text{c}$ is the set with $39$ elements, consisting of the clubs, hearts and spades. For $A$ and $B$ any two measurable sets, their intersection $A \cap B$ is the specification "both $A$ and $B$ occur"; while their union $A \cup B$ is the specification "either $A$ or $B$ (or, possibly, both) occur". In our example, let $A$ represent the specification "is a diamond", and $B$ the specification "is a $9$" (so $B$ is a set with $4$ elements). Then $A \cap B$ is the specification "is a diamond and is a $9$" (a set with just one element, namely the $9$ of diamonds); while $A \cup B$ is the specification "is a diamond or a $9$" (a set with $16$ elements, consisting of the diamonds and the nines).
For $A$ any measurable set, its measure, $\mu(A)$, is interpreted as the probability that, when the experiment is performed, the result will be in the set $A$ of possible outcomes. In our example, for $A$ the set of diamonds, we would demand that $\mu(A) = 1/4$, the probability that a card drawn at random is a diamond. Then $\mu(X) = 1$ is interpreted as "the probability of some outcome is one"; $\mu(\emptyset) = 0$ as "the probability of no outcome is zero"; and that $\mu(A) \le 1$ for all measurable $A$ as "probabilities cannot exceed one". Next, let $A_1, A_2, \ldots$ be disjoint measurable sets ("the outcomes described by the $A_i$ are mutually exclusive"), and consider their union, $\cup A_i$ ("the outcome is either that specified by $A_1$, or that specified by $A_2$, etc."). Then the mathematical fact that$$\mu(\cup A_i) = \sum \mu(A_i)$$has the following interpretation: "The probability that the outcome is one of the mutually exclusive outcomes $A_i$ is the sum of the probabilities for the individual outcomes $A_i$." In our example, let $A$ be the set of diamonds, and $B$ is the set consisting of the $44$ of hearts, the $5$ of spades, and the $4$ of clubs. Then these are disjoint measurable sets. We have in this case$$\mu(A) = 1/4, \text{ } \mu(B) = 3/52, \text{ and } \mu(A \cup B) = 4/13,$$and indeed,$$1/4 + 3/52 = 4/13,$$as required by the general property of probability measures. As a final example, if $A$ and $W$ are measurable sets, with $\mu(W) \neq 0$, then $\mu(A \cap W)/\mu(W)$ is interpreted as the probability of $A$, given that $W$ has occurred. In our example, if $W$ is the set of diamonds, and $A$ the set consisting of the $3$ of diamonds, then $5$ of hearts and the $8$ of spades, then$$\mu(A \cap W)/\mu(W) = (1/52)/(1/4) = 1/13$$the probability that a card drawn, given that it is a diamond, is one of the $3$ diamonds, the $5$ of hearts or the $8$ of spades.
