# Fundamental Theorem of Calculus for Riemann and Lebesgue

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

• You should take care that as stated, the theorems have hypotheses of different natures. For Riemann, you wanted $f'$ to be continuous (actually, this is unnecessary; it need only be integrable). For Lebesgue, you wanted $f$ itself to be absolutely continuous. – Ryan Reich Mar 19 '14 at 16:38
• @Ryan Reich Do you know why it is usually stated in the stronger form of $f′$ needing to be continuous on $[a,b]$? – user100431 Mar 19 '14 at 17:04
• Probably because it's easier than stating what a Riemann-integrable function is (if I recall, Riemann integrable is equivalent to bounded and almost everywhere continuous). – Ryan Reich Mar 19 '14 at 17:37

## 1 Answer

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.

• Thanks for your response. Would I be right in stating that since $f′$ exists only almost everywhere that it does not exist at at most countably many points in $[a,b]$? – user100431 Mar 19 '14 at 16:55
• No, at least not without proof. There are uncountable sets with lebesgue measure zero! Whether or not such sets can arise as the set of non-differentiable points of absolutely continuous function, I don't know, but my guess would be they can. You might want to search this site to see if anybody ever asked for an example for such a function, and if the search comes up empty you might want to post a question about it... – fgp Mar 19 '14 at 17:23
• I think that the last sentence should be clarified: you can change a function from integrable to non-integrable this way, but you can't keep it integrable and change the value of the integral. – Toby Bartels Nov 30 '16 at 16:54