Why $\lim_{n \to \infty} n\int_{x}^{x+\frac{1}{n}} F(u)du - n\int_{a}^{a+\frac{1}{n}} F(u)du = F(x)-F(a)$ In a proof I'm reading, there's the following statement:
$$\lim_{n \to \infty} n\int_{x}^{x+\frac{1}{n}} F(u)du - n\int_{a}^{a+\frac{1}{n}} F(u)du = F(x)-F(a)$$
where $$F(x) := \int_a^x f(u)du$$
and $f$ is bounded and $L-B$ (Lebesgue-Borel) measurable.
I cannot see how the author reached this conclusion.
I have tried splitting the first integral to be written as such:
$$ n\int_{x}^{x+\frac{1}{n}} F(u)du = n\int_{a}^{x+\frac{1}{n}} F(u)du - n\int_{a}^{x} F(u)du$$
which, at the limit, is actually $F'(x)$, which doesn't seem to get me any closer to obtaining the equation.
Another attempt is observing that this last split equals:
$$n \Big( \int_{a}^{x+\frac{1}{n}} F(u)du - \int_{a}^{x} F(u)du \Big ) $$
which at the limit gives:
$$\lim_{n\to \infty} n \cdot \Big( \lim_{n\to \infty} \int_{a}^{x+\frac{1}{n}} F(u)du - \int_{a}^{x} F(u)du \Big ) $$
$$= \lim_{n\to \infty} n \cdot \Big( \int_{a}^{x} F(u)du - \int_{a}^{x} F(u)du \Big )  = 0$$
which also must be a mistake.
Any advice would be much appreciated.
 A: Seven comments so far, time to get explicit with the assistance.
You want to prove this fact about this sequence:
$$\lim_{n \to \infty} n\int_{c}^{c+\frac{1}{n}} g(x)\,dx  = g(c) \tag{*}$$
assuming only that $g$ is continuous   on some interval $[c,c+1]$.
Your attempts to prove the bigger statment led you to miss this simpler reason why it all works.  Your second edit with two separate limits as $n\to\infty$?  What can I say? Don't do that!
Idea 1.  The mean value theorem for integrals [see comment by @Guangliang] allows you to  claim that for each $n$ there is a point $c_n$ so that $c\leq c_n \leq c + \frac1n$
$$ \int_{c}^{c+\frac{1}{n}} g(x)\,dx  = g(c_n)([c+\frac1n]- c) = \frac1n g(c_n).$$
Since $c_n\to c$ and $g$ is continuous, you know that $g(c_n)\to g(c)$ and you can establish (*).
Idea 2.  Forgot the mean-value theorem?  Or bigger fan of the fundamental theorem?
Set it up this way:  $G(t) = \int_c^t g(x)\,dx$ and hence $G'(c)=g(c)$ because of continuity of $g$.
$$ n\int_{c}^{c+\frac{1}{n}} g(x)\,dx = n[G(c+\frac1n)-G(c)] = \frac{G(c+\frac1n)-G(c)}{\frac1n} \to G'(c)$$
as $n\to\infty$ by the definition of derivative and again   you can establish (*).
Note.  The first method, using the mean-value theorem, is available here since $g$ is continuous.  The seond idea does not need continuity and depends only on $g$ being   integrable in some sense on $[c,c+1]$ and on $g$ being continuous at the point $c$.
Postscript.  By the way, students often stare hard and confused at the limit
$$ \lim_{n\to \infty}  \frac{G(x+\frac1n)-G(x)}{\frac1n}  .$$
Ideally you glance briefly and recognize immediately that must be $G'(x)$ if that derivative exists.  It allows you to see easily that if $G$ is differentiable then $g(x)=G'(x)$ is the pointwise limit of a sequence of continuous functions, i.e., derivatives belong to the first class of Baire.  That is an important first step in understanding derivatives.
