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 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"?

• +1 @Viru, Ahhh I didn't know that, thanks for the help Apr 21, 2020 at 15:04

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.

$$\sum_{k=2}^{\infty}\left(\frac{1}{k-1}\:-\:\frac{1}{k}\right) = 1$$

This sum is simply equal to $$1$$.

• Haven't you proved with this that the width of the interval of integration is $1$? How is possible to know if it is $[0,1]$ and not, for example, $[-3,-2]$? Apr 21, 2020 at 15:36

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.