Is the following inequality correct? How to prove it? Suppose $X$ and $Y$ are two arbitrary random variables, and we have the following inequality that conditional on $Y=y$,
$$\textbf{Pr}(X \ge a_0 | Y=y)\le f(y),$$ 
where $\textbf{Pr}(\cdot)$ denotes the probability of the event, $a_0$ is a constant, and $f(y)$ is an increasing function with respect to $y$. I want to know whether the following inequality is correct,
$$\textbf{Pr}(X \ge a_0 , Y\le b_0)\le f(b_0),$$
where $b_0$ is a constant.
If it is wrong, is there any counter example? Thanks a lot.
 A: Expand using product rule:
$$P(X \ge a_0 \cap Y \le b_0) = P(X \ge a_0 | Y \le b_0) P(Y \le b_0)$$
Assuming $P(Y \le b_0)$ is nonzero, you can use the first inequality to obtain:
$$P(X \ge a_0 | Y \le b_0) P(Y \le b_0) \le f(y\le b_0) P(Y \le b_0)$$
Because $f(y)$ is increasing in $y$, $f(y \le b_0) \le f(b_0)$ so the inequality of interest holds.
By the way, in the context of probability $f$ is normally used for PDFs so you may want to use a different letter to avoid confusion.
EDIT: Granted, $f(y \le b_0)$ is an abuse of notation. It should be read as $f(y), y \le b_0$ and not as the function taking a set for its argument.
A: First, we have
$$\eqalign{
\textbf{Pr}(X \ge a_0|Y = y) &=\int_{a_0}^{+\infty}p_{X}(x|Y= y)dx \\
&=\int_{a_0}^{+\infty}\frac{p_{X,Y}(x,y)}{p_Y(y)}dx\\
&=\frac{1}{p_Y(y)}\int_{a_0}^{+\infty}{p_{X,Y}(x,y)}dx,
}$$
where $p_{X,Y}(x, y)$ is the joint PDF of $(X,Y)$, $p_{X}(x| Y=y)$ is the  PDF of $X$ conditional on $Y=y$, and $p_Y(y)$ is the PDF of $Y$. 
Because when $y \le b_0$, $f(y)\le f(b_0)$, that is 
$$\int_{a_0}^{+\infty}{p_{X,Y}(x,y)}dx \le f(b_0)p_Y(y)  $$
Finally, substitute it into the following,
$$\eqalign{
\textbf{Pr}(X \ge a_0, Y \le b_0) &= \int_{a_0}^{+\infty}\int_{-\infty}^{b_0}p_{X,Y}(x, y)dy dx \\
&\le \int_{-\infty}^{b_0} f(b_0)p_Y(y) dy \\
& \le f(b_0),
}$$ 
which proves the required inequality.
