why $\int_0^1f= \lim_{n\to+\infty}\sum_{k=2}^n\int_{\frac{1}{k}}^{\frac{1}{k-1}}f$ I'm looking for an explanation for why this equality is true assuming f is integrable. obviously it is true for a finite amount of integrals covering exactly [a,b], but how do we know it is also true for the infinite case?
My intuition is that the part of $[0,1]$ we're ignoring for each n gets smaller, meaning it converges to 0, and therefore both sides are identical. Am i right thinking this way?
If anyone cares to write a proof to convince me it will be greatly appreciated.
$$\int_0^1f= \lim_{n\to+\infty}\sum_{k=2}^n\int_{\frac{1}{k}}^{\frac{1}{k-1}}f$$
 A: let partial sum be
$$S_n = \sum_{k=2}^n \int_{\frac{1}{k}}^{\frac{1}{k-1}}f$$ then we have to show that $\lim_{n \rightarrow \infty} S_n = \int_0^1 f$ and key point we will use is that $ \exists M>0 \,\,\, |f| < M$ since $f$ is Riemann Integrable (i.e it is bounded)
$$ \Big|\sum_{k=2}^n \int_{\frac{1}{k}}^{\frac{1}{k-1}} f - \int_{0}^1 f \Big| = |\int_{1/n}^1 f - \int_{0}^1f | = \Big|\int_0^{\frac{1}{n}} f \Big| \leq \int_0^{\frac{1}{n}} |f| \leq \frac{M}{n} $$ 
which implies that $n>\frac{M}{\epsilon}$ suffices and then we have limit
Edit:@Aditya proof is also good but existence of antiderivative is not necessary and it is doesn't imply from Integrability  look at examples here : Is "integrability" equivalent to "having antiderivative"?
A: Let us assume that the antiderivative of $f$ is $F$.
By the Fundamental Theorem of Calculus,
$$\therefore \int_{\frac{1}{k}}^{\frac{1}{k-1}}f(x)dx =F(\frac{1}{k-1})-F(\frac{1}{k})$$
$$\therefore \sum_{k=2}^{n}\int_{\frac{1}{k}}^{\frac{1}{k-1}}f(x)dx=\sum_{k=2}^{n}F(\frac{1}{k-1})-F(\frac{1}{k})=F(1)-F(\frac{1}{n-1})$$
$$\therefore \lim_{n \to \infty}F(1)-F(\frac{1}{n-1})=F(1)-F(0) = \int_{0}^{1}f(x)dx$$ Which proves your result. Hope this helps.
A: $$\sum_{k=2}^{\infty}\left(\frac{1}{k-1}\:-\:\frac{1}{k}\right) = 1$$
This sum is simply equal to $1$.
A: The set $(0,1]$ is the disjoint union
$$
(0,1] = \bigcup_{k=2}^{\infty} \left(\frac{1}{k},\frac{1}{k-1}\right]
$$
Therfore, if $f$ is Lebesgue integrable on $(0,1]$ we have
$$
\int_{(0,1]} f = \sum_{k=2}^\infty \int_{(1/k,1/(k-1)]} f
$$
Use the dominated convergence theorem.
