Real-Analysis The Fundamental Theorem of Calculus, guaranteed and verified 100% Pass

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The Fundamental Theorem of Calculus


We saw earlier that if 𝑓 is a continuous function on [𝑎, 𝑏] then
𝑏
∫𝑎 𝐷𝑖𝑓𝑓ℎ 𝑓 = 𝐴𝑣ℎ 𝑓(𝑏) − 𝐴𝑣ℎ 𝑓(𝑎)
𝑓(𝑥+ℎ)−𝑓(𝑥) 1 𝑥+ℎ
where 𝐷𝑖𝑓𝑓ℎ 𝑓 (𝑥 ) = and 𝐴𝑣ℎ 𝑓(𝑥 ) = ∫𝑥 𝑓.
ℎ ℎ



Since 𝑓 is continuous:
𝑏 𝑏
lim+ ∫𝑎 𝐷𝑖𝑓𝑓ℎ 𝑓 = ∫𝑎 𝑓′ if the limit exists and
ℎ→0

lim (𝐴𝑣ℎ 𝑓(𝑏) − 𝐴𝑣ℎ 𝑓(𝑎)) = 𝑓(𝑏) − 𝑓 (𝑎).
ℎ→0+




So we get the fundamental theorem of Calculus:
𝑏
∫𝑎 𝑓′ = 𝑓(𝑏) − 𝑓(𝑎).



The question is, is this statement still true even if 𝑓(𝑥) is not differentiable
everywhere on (𝑎, 𝑏)? If not, when is it true?

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Theorem: Let 𝑓 be absolutely continuous on the closed, bounded interval [𝑎, 𝑏].
Then 𝑓 is differentiable a.e. on [𝑎, 𝑏] and
𝑏
∫𝑎 𝑓′ = 𝑓(𝑏) − 𝑓(𝑎).


Proof: Since 𝑓 is absolutely continuous, it is the difference of two increasing
absolutely continuous function on [𝑎, 𝑏].


Therefore, by Lebesgue’s theorem 𝑓 is differentiable a.e. on (𝑎, 𝑏).

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𝑓(𝑥+𝑛)−𝑓(𝑥)
Thus {𝐷𝑖𝑓𝑓 1 𝑓} = { 1 } converges pointwise a.e. on (𝑎, 𝑏) to 𝑓′.
𝑛 𝑛



In addition, since 𝑓 is absolutely continuous, {𝐷𝑖𝑓𝑓1 𝑓} is uniformly integrable
𝑛
over [𝑎, 𝑏].


Since {𝐷𝑖𝑓𝑓1 𝑓} is uniformly integrable over [𝑎, 𝑏] and 𝐷𝑖𝑓𝑓 1 𝑓 → 𝑓 ′
𝑛 𝑛
pointwise a.e. on (𝑎, 𝑏), the Vitali Convergence Theorem says 𝑓′ is integrable and
𝑏 𝑏 𝑏
lim ∫𝑎 𝐷𝑖𝑓𝑓1 𝑓 = ∫𝑎 lim 𝐷𝑖𝑓𝑓1 𝑓 = ∫𝑎 𝑓′.
𝑛→∞ 𝑛 𝑛→∞ 𝑛


From first year Calculus we know that if 𝑓 is continuous then:
𝑏
lim (𝐴𝑣1 𝑓(𝑏) − 𝐴𝑣 1 𝑓(𝑎)) = lim ∫ 𝐷𝑖𝑓𝑓 1 𝑓
𝑛→∞ 𝑛 𝑛 𝑛→∞ 𝑎 𝑛
𝑏
𝑓(𝑏) − 𝑓(𝑎) = ∫𝑎 𝑓′.