A question regarding an integrable function This question arose while I was reading a paper. 
Let $f$ be a positive real valued function that is integrable on $\mathbb{R}$. So,there exists $F > 0$ s.t
$$\int_{\mathbb{R}} f(x)dx = F~.$$
Now, the author claims that it's possible to find a function $u$ : $(0,1) \to \mathbb{R}$ s.t $$\dfrac{1}{F} \int_{-\infty}^{u(t)} f(x)dx = t~.$$
Here comes the tricky part,
The author claims that, $u$ may be discontinuous but it's strictly increasing ;hence it's differentiable almost everywhere.

The latter part of the statement follows from the Lebesgue's Differentiation theorem. But,
How does the author claim that $u$ must be strictly increasing? Is it possible for $u$ to be discontinuous? (It's intuitive to think that the existence of such $u$ is possible but I'm interested in a rigorous proof of this claim)

My thoughts,
Since $f$ is integrable, let $f_0$ be it's anti-derivative so,$$\dfrac{1}{F} \int_{-\infty}^{u(t)} f(x)dx = \dfrac{1}{F} [f_0 (x)]_{-\infty}^{u(t)} = t~.$$
$$\dfrac{1}{F}(f_0 (u(t)) - \lim_{x \to -\infty} f_0(x)) = t$$
From this, it can be concluded that $f_0$ increases as $t$ increases. But, can we conclude $u$ increases as well?
 A: If $u(s)=u(t)$, then
$$
s = \frac{1}{F} \int_{-\infty}^{u(s)} f(x)\,dx = \frac{1}{F} \int_{-\infty}^{u(t)} f(x)\,dx = t;
$$
therefore $u$ is injective. Coupled with the fact that $u$ is increasing, this shows that $u$ is strictly increasing.
If $f$ is strictly positive, then $u$ is in fact continuous. The only discontinuity a strictly increasing function can have is a jump discontinuity; but if $u(t-) = a$ and $u(t+)=b>a$, then
\begin{align*}
0 < \frac{1}{F} \int_a^b f(x)\,dx &= \frac{1}{F} \int_{-\infty}^b f(x)\,dx - \frac{1}{F} \int_{-\infty}^a f(x)\,dx \\
&= \frac{1}{F} \int_{-\infty}^{u(t+)} f(x)\,dx - \frac{1}{F} \int_{-\infty}^{u(t-)} f(x)\,dx \\
&= (t+)-(t-) = 0,
\end{align*}
a contradiction.
If $f$ is merely nonnegative, then $u$ can be discontinuous: when $f$ is the indicator function of the set $[0,\frac12]\cup[\frac32,2]$, then
$$
u(t) = \begin{cases}
t, &\text{if } 0<t<\frac12, \\
t+1, &\text{if } \frac12<t<1.
\end{cases}
$$
(The value $u(\frac12)$ can be anywhere between $\frac12$ and $\frac32$.)
A: When $f>0$ the value of the integral is increasing in the upper limit of integration. When $u$ is differentiable, applying Leibniz's integral rule to your second equation yield 
$$ f(u(t)) u'(t) = 1 \Rightarrow u'(t)>0. $$
