Properties of inverse distribution function - continuity I have some questions about inverse distribution functions. Let $F : \mathbb R \to [0,1]$ be a distribution function and define $F^{-1} : [0,1] \to \overline{\mathbb R}$ by $F^{-1}(y) := \inf\{x \in \mathbb R; F(x) \ge y\}$ with the convention that $\inf \emptyset := +\infty$. Futhermore, define $F^{-1+} : [0, 1] \to \overline{\mathbb R}$ by $F^{-1+}(y) := \sup\{x \in \mathbb R; F(x) \le y\}$, where $\sup \emptyset := -\infty$.
The function $F^{-1}$ is the usual quantile function. I managed to prove the following properties:


*

*$F^{-1}$ and $F^{-1+}$ are non-decreasing.

*If $F^{-1} \in (-\infty, \infty)$, $F^{-1}$ is left-continuous at $y$ and admit a limit from the right at $y$.

*For $y \in \mathbb R$, set $A_y := \{x \in \mathbb R; F(x) = y\}$. Then, $(F^{-1}(y-), F^{-1}(y+)) \subseteq A_y$.

*Suppose that $F$ is strictly increasing. Then, $F^{-1}$ is continuous.


It would be nice, if I could prove the properties 2 and 4 also for $F^{-1+}$. Do they hold also for $F^{-1+}$? I needed property 3 to prove 4.
 A: The function $F^{-1+}$ is continuous from the right. To see this let
$y_{0}$ be such that $x_{0}:=F^{-1+}(y_{0})\in\mathbb{R}$ and consider a
sequence $y_{n}\searrow y_{0}$. Set $F^{-1+}(y_{n}):=x_{n}$. Since $F^{-1+}$
is non-decreasing, $F^{-1+}(y_{0})\leq F^{-1+}(y_{n+1})\leq F^{-1+}(y_{n})$
and so $x_{0}\leq x_{n+1}\leq x_{n}$. It follows that $x_{n}\searrow x$ with
$x_{0}\leq x$. 
Since $x_{n}=F^{-1+}(y_{n})=\sup\{x:\,F(x)\leq y_{n}\}$ for every
$\varepsilon>0$ we have $F(x_{n}-\varepsilon)\leq y_{n}<F(x_{n}+\varepsilon)$.
If by contradiction $x_{0}<x$, then taking $\varepsilon:=\frac{x-x_{0}}{2}$ we
have that $x_{n}-\varepsilon>x_{0}+\varepsilon$ and so by monotonicity
$y_{n}\geq F(x_{n}-\varepsilon)\geq F(x_{0}+\varepsilon)>y_{0}$. Letting
$n\rightarrow\infty$ and using the fact that $y_{n}\searrow y_{0}$ we get a
contradiction. This shows that $x_{n}\searrow x_{0}$, that is, that $F^{-1+}$
is continuous from the right. 
That $F^{-1+}$ admits a limit from the left follows from the fact that
$F^{-1+}$ is non-decreasing. So (2) holds with right continuous in place of left.
A: So $F_X(x)=P(X\le x)$
I will prove (4.) - "Suppose that $F$ is strictly increasing. Then, $F^{−1+}$ is continuous"
Opposite would be: there is $y_o$ and there is $m>0$ such that $\forall\delta>0,\exists y,|y-y_0|<\delta \land |F^{−1+}(y)-F^{−1+}(y_0)|>m$ 
Let $x_0=F^{−1+}(y_0)$ since $F$ is right continuous then $F(x_0)\ge y_0$
So we have 
$$F(x_0-m)<y_1<F(x_0-m/2)\le y_0\le F(x_0)<y_2<F(x_0+m/2)$$
where $y_1$ and $y_2$ are arbitrary numbers that satisfy the inequatity.
Also analogically 
$$x_0-m\le F^{−1+}(y_1)\le x_0\le F^{−1+}(y_2)\le x_0+m/2$$
and  since $F^{−1+}$ is increasing $|F^{−1+}(y)-x_0|<m$ for all {$y|y_1<y<y_2$}
EDIT - further explanation of some statements
Why is $F(x_0) \ge y_0$?
If it was not, meaning $F(x_0) < y_0$, then from definition of $F^{-1+}$ we have $F(x)>y_0$ for all $x>x_0$, in other words $P(x_0<X\le x)>y_0-F(x_0)$ for all $x>x_0$ which is impossible because $F$ has to be right continuous, see Cumulative distribution function and right continuity
A: $F$ is a distribution function, so at a point $x$ this function may only behave in a few different ways:

*

*It may be continuous at $x$ and strictly increasing immediately after $x$. This means for all $x' > x$ we have $F(x') > F(x)$.

*It may be continuous at $x$ and "flat" immediately after $x$. This means there is some $x' > x$ for which $F(x') = F(x)$.

*It may have a jump discontinuity at $x$.

You can see each of these behaviors in the following picture:

You can also "see" (and then prove, as in the other answers) that the only values of $y$ where $F^{-1+}(y)$ is discontinuous are the heights of the "flat parts" of $F$, in which case $F^{-1+}$ makes a right-continuous jump. In light of this observation, your (2) is not true for $F^{-1+}$, which is right-continuous at its jumps, but your (4) is true, since a strictly increasing $F$ has no flat parts.
