$F\big( g(t) \big) - F\big( g(t + h) \big) \leq h$ implies that $g$ is right-continuous? Suppose $F$ is a continuous, strictly increasing distribution function. Also,
suppose that $g:[0,1] \longrightarrow [0,1]$ such that for any $t \in [0,1]$,
$h > 0$, and $\epsilon > 0$, 
$$
F\big( g(t) \big) - F\big( g(t + h) \big) \leq h
$$
and
$$
F\big( g(t) \big) - F\big( g(t+h) - \epsilon \big) \geq h - P\big( hf(X) > \epsilon \big)
$$
where $f(\cdot) \geq 0$ is a function such that $E[ f(X) ] < \infty$.
Are these conditions sufficient to show that $g$ is right-continuous?
 A: Choose $t\in[0,1)$ and assume, for any chosen $t$, $\lim\limits_{h\downarrow 0}g(t+h)$ exists. By the boundedness of $g$, these limits lie between $0$ and $1$. 
Now,  using the continuity of $F$ on $[0,1]$, note that,
$$
\lim\limits_{h\downarrow 0}F\big(g(t)\big)-\lim\limits_{h\downarrow 0}F\big(g(t+h)\big) = F\big(g(t)\big)-F\big(\lim\limits_{h\downarrow 0}g(t+h)\big) \leqslant \lim\limits_{h\downarrow 0}h = 0\,.
$$
Therefore,

$$
F\big(g(t)\big) \leqslant F\big(\lim\limits_{h\downarrow 0}g(t+h)\big)\,.\tag{1}
$$

Alternatively, since (for $\epsilon>0$)
$$
P\big( hf(X) > \epsilon \big) = \mathbb{E}\left[{\bf 1}_{\{hf(X)>\epsilon\}}\right] \leqslant  \mathbb{E}\left[\frac{h}{\epsilon}f(X){\bf 1}_{\{hf(X)>\epsilon\}}\right] \leqslant  \frac{h}{\epsilon}\mathbb{E}\left[f(X)\right],
$$
we have,
$$
\lim\limits_{h\downarrow 0}F\big( g(t) \big) - \lim\limits_{h\downarrow 0}F\big( g(t+h) - \epsilon \big) \geqslant \lim\limits_{h\downarrow 0}h - \lim\limits_{h\downarrow 0}P\big( hf(X) > \epsilon \big) \geqslant \lim\limits_{h\downarrow 0}h - \lim\limits_{h\downarrow 0}\frac{h}{\epsilon}\mathbb{E}\left[f(X)\right]) = 0\,.
$$
So, for each $\epsilon > 0$ small enough,
$$
F\big(g(t)\big) \geqslant \lim\limits_{h\downarrow 0}F\big( g(t+h) - \epsilon \big) = F\big( \lim\limits_{h\downarrow 0}g(t+h) - \lim\limits_{h\downarrow 0}\epsilon \big) = F\big( \lim\limits_{h\downarrow 0}g(t+h) - \epsilon \big)\,.
$$
Therefore,

$$
F\big(g(t)\big) \geqslant F\big( \lim\limits_{h\downarrow 0}g(t+h) \big)\,.\tag{2}
$$

Together, $(1)$ and $(2)$ imply
$$
F\big(g(t)\big) = F\big( \lim\limits_{h\downarrow 0}g(t+h) \big)\,.
$$ 
But, over $[1,0]$, $F$ is one-to-one, hence

$$
g(t) = \lim\limits_{h\downarrow 0}g(t+h)\,,\quad\forall\ t\in[0,1)\,.
$$ 

