Prove or disprove: If X has a cdf F then P(F(X) ≤ a) ≤ a. Under what condition on F will you get P(F(X) ≤ a) = a? Prove or disprove: If X has a cdf F then P(F(X) ≤ a) ≤ a. Under what condition on F will you get P(F(X) ≤ a) = a?
I think if a is negative, then the statement is false. Any hint for the equal condition? 
 A: Implicitly, you introduced a new random variable, $Y = F(X)$. By definition, $Y \in [0, 1]$. 
Suppose that $F(X)$ is differentiable and strictly increasing. Suppose also that $F(X)$ is invertible. Let's $f(X) = F'(X)$ be the pdf of $X$. Then the pdf of $Y$ is:
$$f_Y(Y) = \displaystyle f(\bar{X})\frac{d \bar{X}}{d Y},$$
where $Y = F(\bar{X})$ or equivalently $\bar{X} = F^{-1}(Y)$. Then:
$$f_Y(Y) = \displaystyle f(F^{-1}(Y))\frac{d F^{-1}(Y)}{d Y}.$$
Recall that:
$$\frac{d F^{-1}(Y)}{d Y} = \frac{1}{f(F^{-1}(Y))},$$
and hence $f_Y(Y) = 1 ~\forall ~Y \in [0,1]$ and $0$ elsewhere. This means that $Y$ is a uniformly distributed random variable in the set $[0,1]$ whichever is the distribution of $X$.
Finally:
$$P(F(X) \leq a) = P(Y \leq a) = \begin{cases}
0 & \text{if $a < 0$} \\
a & \text{if $a \in [0, 1]$} \\
1 & \text{if $a > 1$}
\end{cases}
$$
Then $$P(F(X) \leq a) \leq a$$ if $a \in [0, 1]$. 
A: The only relevant values of $a$ are in $[0,1)$.
If $F$ is continuous, then $\{F(X)\le a\}=\{X\le G(a)\}$, where $G$ is the right-continuous inverse of $F$ defined by
$$
G(y):=\inf\{x:F(x)>y\},\qquad 0\le y<1.
$$
In this case, $P[F(X)\le a]=P[X\le G(a)]=F(G(a))=a$, the final equality resulting from the right continuity of $F$.
In general, we only have the inclusion $\{F(X)\le a\}\subset\{X\le G(a)\}$, yielding $P[F(X)\le a]\le P[X\le G(a)]=F(G(a))=a$, for $a\in[0,1)$.
Example: $X$ takes the values $0$ and $1$ with probabilities $1/2$ each. Let $a=1/4$. Then $P[F(X)\le 1/4]=P[F(X)=0]=0$ but $G(1/4)=1/2$ and $P[F(X)\le 1/2]=P[X\le 0]=1/2$.
