Showing $F(x) = 1/x \int_0^x f(t) dt$ is increasing In proving : 
If $f$ is bounded variation on $[0,1]$, then $$F(x) = (1/x) \int_0^x f(t) dt$$ is of bounded variation on $[0,1]$.
I am attempting to prove:  If $f$ is increasing, then $F$ is increasing.
I attempted to bypass the problem in numerous ways but I was unsuccessful.
If I have the claim, I solve the problem.
Can anyone give me a hint?
Any help would be appreciated!
 A: We want to show that $x < y \Rightarrow F(x) \leq F(y)$, i.e. $$\frac{1}{x}\int_0^x f(t)\,\mathrm{d}t \leq \frac{1}{y}\int_0^y f(t)\,\mathrm{d}t$$ We note that $$\frac{1}{y}\int_0^y f(t)\,\mathrm{d}t = \frac{1}{y}\left[\int_0^x f(t)\,\mathrm{d}t+\int_x^y f(t)\,\mathrm{d}t\right]$$ Therefore, we can rewrite our inequality as $$\left(\frac{y}{x}-1\right)\int_0^x f(t)\,\mathrm{d}t \leq \int_x^y f(t)\,\mathrm{d}t$$ We multiply both sides by $\frac{1}{y-x}$ to get $$\frac{\frac{y}{x}-1}{y-x}\int_0^x f(t)\,\mathrm{d}t = \frac{1}{x}\int_0^x f(t)\,\mathrm{d}t \leq \frac{1}{y-x}\int_x^y f(t)\,\mathrm{d}t$$ The left-hand side is the average value of $f(t)$ on $(0, x)$, while the right-hand side is the average value of $f(t)$ on $(x, y)$. The left-hand side is therefore bounded above by $\sup_{t\in (0, x)} f(t)$, while the right-hand side is bounded below by $\inf_{t\in (x, y)} f(t)$. As $f$ is increasing, we therefore have that $$\frac{1}{x}\int_0^x f(t)\,\mathrm{d}t \leq \sup_{t\in (0, x)} f(t) \leq \inf_{t\in (x, y)} f(t) \leq \frac{1}{y-x}\int_x^y f(t)\,\mathrm{d}t$$ so we are done.
A: Calculate the derivative of $F$: $F'(x) = \frac{\int_0^x (f(x) - f(t)) dt}{x^2}$ almost everywhere. We see that if $f$ is non-decreasing then $F' \geq 0$.
A: A simple proof, using a linear substitution: For $0 < x < y$ is
$$
\begin{align}
\boxed{F(x)} &= \frac 1x \int_0^x f(t) \, dt \\
&= \frac 1y \int_0^y f(\frac xy s) \,ds \quad \text{(substitution $t = \frac xy s$)} \\
&\boxed{\le} \frac 1y \int_0^y f(s) \, ds \quad \text{(because $f$ is increasing)}\\
 &= \boxed{F(y)} \, .
\end{align}
$$
