Implied expressions with same probability Given two variables $a$ and $b$ and two intervals $I$ and $J$, is the following affirmation true? $$
\Pr[a\in I]\geq p\ \land\ (a\in I\implies b\in J) \implies \Pr[b\in J]\geq p
$$
Of course, $a\in I$ and $b\in J$ both depend on one or more random variables; otherwise, it wouldn't make sense to speak about probabilities.
In case it's not true, I would like to see the proof. I have no idea where to start, and I would like to know also if there's some minimum set of requirements that would make that affirmation true.
 A: This is really a question about sets and mathematical notation, so let's unravel the notation.
Random variables are, first of all, functions.  Abstractly their domains are a sample space $\Omega$ and we don't need to be concerned with their codomains (usually, but not always, the set of real numbers).  Thus we may more formally and explicitly write
$$(a,b):\Omega \to \mathcal{X}_a \times \mathcal{X}_b$$
where the codomain of the bivariate random variable $(a,b)$ is the Cartesian product of their individual codomains.
In this context, $I \subseteq \mathcal{X}_a$ and $J\subseteq\mathcal{X}_b$ are certain subsets of these codomains.  The conditions potentially constrain the possible values of $a$ and $b:$

*

*$a\in I$ refers to that set of observations $\omega\in\Omega$ for which $a(\omega)\in I,$ written $a^{-1}(I).$


*$b\in J$ refers to that set of observations $\omega\in\Omega$ for which $b(\omega)\in J,$ written $b^{-1}(J).$
Therefore, "$a\in I\implies b\in J$" means that if $a(\omega)\in I$ then $b(\omega)\in J.$  See the figure.

Equivalently, this can be expressed as
$$\text{For all }\omega\in\Omega,\ \omega \in a^{-1}(I)\implies \omega \in b^{-1}(J).$$  But this is the very definition of set inclusion, whence
$$a^{-1}(I) \subseteq b^{-1}(J).$$
The monotone axiom of probability says the probability of a subset is no greater than the probability of any set in which it is contained.  In particular, combining this with the assumptions in the question, we see
$$p \le \Pr(a \in I) = \Pr(a^{-1}(I)) \le \Pr(b^{-1}(J)) = \Pr(b\in J).$$
The first "$\le$" is an assumption in the question; the "$=$" are the notation explained above; and the middle "$\le$" arises from applying the monotone axiom.
