Lebesgue integral limit Suppose $f:\mathbb{R}^+ \to \mathbb{R}$ is Lebesgue integrable on every finite interval and 
$$I(x;\alpha) = \int_\alpha^{\alpha+x} f(t) dt$$
 If $\lim_{\alpha \to \infty}I(x; \alpha) = F(x)$ exists for every $x$, then show $F(x) = Cx$ for some real number $C$.
My attempt is to first consider a continuous function with a finite limit at infinity.  Then by mean value theorem there exists a point $\xi_\alpha > \alpha$ where
$$\lim_{\alpha \to \infty}\int_\alpha^{\alpha+x} f(t) dt = \lim_{\alpha \to \infty}f(\xi_\alpha)x = [\lim_{t\to \infty}f(t)]x $$
I do not see how to prove the result without these strong assumptions.
 A: Note that
$$\int_\alpha^{\alpha + x_1 +x_2} f(t) \, dt = \int_\alpha^{\alpha + x_1} f(t) \, dt + \int_{\alpha+x_1}^{(\alpha + x_1)+ x_2} f(t) \, dt$$
Since $\alpha + x_1 \to \infty$ iff $\alpha \to \infty$ it follows that $F$ is additive:
$F(x_1+x_2) = F(x_1) + F(x_2)$.
With any sequence $\alpha_n \to \infty$, we have $F(x) = \lim_{n \to \infty}I(x;\alpha_n)$ and $F(x)$ is a pointwise limit of a sequence of continuous functions. Using Baire's theorem, it can be shown that $F$ is continuous on an everywhere dense set.
However, we only need continuity at one point to conclude that $F$, as an  additive function, is everywhere continuous with $F(x) = F(1)\cdot x$.
Addendum 
To elaborate, it is easy to prove that an additive function has the property $F(rx) = rF(x)$ for any rational number $r$.  Hence, for any real number $\xi$, taking a sequence of rationals $r_n \to \xi$, we have if $F$ is continuous,
$$\xi F(x) = \lim_{n \to \infty}r_nF(x) = \lim_{n \to \infty}F(r_n x) = F(\xi x)$$
Also, if $F$ is additive and continuous at one point, say $c$, then it is continuous everywhere since
$$F(x + \delta) - F(x) = F(\delta) = F(c + \delta) - F(c)$$
That $F$ must be continuous at one point is a somewhat deeper result requiring a lengthy argument. For example, see here.
