Fundamental Theorem of Calculus for Riemann and Lebesgue

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Quick question regarding the second part of the Fundamental Theroem of Calculus in terms of Riemann and Lebesgue Integration:

In terms of applying the second part of fundamental theorem of calculus, we have that when considering Riemann integration the condition necessary is $f'$ exists and is continuous in $[a,b]$ implies that we can write $\int_{[a,b]}f'(x)dx = f(b) - f(a)$.

For Lebesgue integration we have $f$ is absolutely continuous in $[a,b]$ implies if that we can write $\int_{[a,b]}f'(x)dx = f(b) - f(a)$.

I can also that the Riemann condition implies the Lebesgue condition. In other words, $f'$ exists and is continuous in $[a,b]$ implies that $f$ is absolutely continuous on $[a,b]$.

My question is why can we not necessarily write "$\int_{[a,b]}f'(x)dx = f(b) - f(a)$" for Riemann integration if $f$ is absolutely continuous only? Why do we need a stronger condition?

Thanks

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Not every absolutely continuous function $f$ is differentiable, so it's not self-evident what $$ \int_{[a,b]} f'(x) dx $$ even means for a general absolutely continuous $f$.

For lebesgue integration, the validity of your statement hinges on the fact that absolute continuity implies differentiability almost everywhere, and that changing a function almost nowhere (i.e., on a set of measure zero) won't change the integral. Thus, it's doesn't matter what values you assume $f'(x)$ takes for those $x$ where $f$ isn't differentiable, because the integral will be the same no matter what.

For riemann integration, however, even changing a function at only coutably many points can change the value of the integral, so this doesn't work.