An alternative proof for the first part involves the use of a formula derived by the method of exhaustion: $$ \int_{a}^{b}f(t)dt=\left( b-a\right) \sum_{n=1}^{\infty}\sum_{m=1}^{2^{n} -1}\left( -1\right) ^{m+1}2^{-n}f\left( a+m(b-a)/2^{n}\right) . $$

Given that $$ F(x)=\int_{a}^{x}f(t)dt, $$

and $$ F^{\prime}(x)=\lim_{\Delta x\rightarrow0}\frac{F(x+\Delta x)-F(x)}{\Delta x}=\lim_{\Delta x\rightarrow0}\frac{1}{\Delta x}\int_{x}^{x+\Delta x}f(t)dt, $$

the above formula leads to: $$ F^{\prime}(x)=\lim_{\Delta x\rightarrow0}\frac{(x+\Delta x-x)}{\Delta x} \sum_{n=1}^{\infty}\sum_{m=1}^{2^{n}-1}\left( -1\right) ^{m+1}2^{-n}f\left( x+m\Delta x/2^{n}\right) , $$

or $$ F^{\prime}(x)=\sum_{n=1}^{\infty}\sum_{m=1}^{2^{n}-1}\left( -1\right) ^{m+1}2^{-n}f\left( x\right) . $$

Since it can be shown that $$ \sum_{n=1}^{\infty}\sum_{m=1}^{2^{n}-1}\left( -1\right) ^{m+1}2^{-n} =\sum_{n=1}^{\infty}2^{-n}=1, $$

It follows that $$ F^{\prime}(x)=f(x). $$

The second part of the proof is identical to the parent.