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.
$$\int_0^1f= \lim_{n\to+\infty}\sum_{k=2}^n\int_{\frac{1}{k}}^{\frac{1}{k-1}}f$$

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.
