The Fundamental Theorem of Calculus

Rigorous Analysis with the Lebesgue Integral

Most people are familiar with the inverse relationship between the integral operation and the derivative, referred to as The Fundamental Theorem of Calculus. We are typically introduced to this relationship in a first term calculus course. I will provide a more rigorous outline of this relationship, with the Lebesgue integral, and show that it (under certain circumstances) still obeys this relationship. Really there are two statements to consider. I will state them both and then resolve them in order.

What to Consider

The first question: does integrability on an interval imply the existence of a derivative? Additionally, does the derivative of the integrated function return the unintegrated function? More formally:

If \(f\) is an integrable function on \([a,b]\) and \(F(x) = \int_a^x f(y)\,dy\), can we conclude that \(F'(x)\) exists, and that \(F'(x)=f(x)\) (at least for a.e. \(x\))?

The second question flips this: what restrictions on a function \(F\) defined on \([a,b]\) guarantee that the derivative exists almost everywhere, is integrable, and that

F(b)-F(a)=\int_a^b F'(x)\,dx?

The Derivative of an Integral

We define:

F(x)=\int_a^x f(y)\,dy, \qquad a\le x\le b.

To see if the derivative exists, we use the definition:

F'(x)=\lim_{h\to 0}\frac{F(x+h)-F(x)}{h} = \lim_{h\to 0}\frac{1}{h}\int_x^{x+h} f(y)\,dy.

Interpreting the interval \([x,x+h]\) as a “ball” \(B\) with measure \(m(B)=h\), we can rewrite this as the averaging problem: does the average value over shrinking neighborhoods converge to \(f(x)\)?

The Lebesgue Differentiation Theorem

Theorem: If \(f\in L^1(\mathbb{R}^d)\), then

\lim_{m(B)\to 0}\frac{1}{m(B)}\int_B f(y)\,dy = f(x)

for almost every \(x\).

Thus, the derivative of the integral recovers the original function almost everywhere, provided \(f\) is integrable. Behavior at infinity does not affect averages on finite balls, meaning local integrability (\(f\in L^1_{\text{loc}}\)) suffices.

The Lebesgue Set

For \(f\in L^1_{\text{loc}}\), the Lebesgue set of \(f\) is the set of points \(x\) such that:

\lim_{m(B)\to 0}\frac{1}{m(B)}\int_B |f(y)-f(x)|\,dy = 0.

Almost every point in \(\mathbb{R}^d\) belongs to the Lebesgue set of \(f\).

The Integral of a Differentiated Function

We now consider the second direction. Not all continuous functions are differentiable everywhere (e.g., the Weierstrass function). We need specific conditions to ensure \(F(b)-F(a)=\int_a^b F'(x)\,dx\).

Bounded Variation vs. Absolute Continuity

A function has Bounded Variation if the sum of its oscillations is finite. While such functions are differentiable a.e., they may still fail the FTC (like the Cantor "Devil's Staircase" function).

Absolute Continuity: \(F\) is absolutely continuous on \([a,b]\) if for every \(\varepsilon>0\) there exists \(\delta>0\) such that

\sum (b_k-a_k)<\delta \;\Rightarrow\; \sum |F(b_k)-F(a_k)|<\varepsilon

for any finite disjoint collection of intervals.

Theorem (Integral of the Derivative): If \(F\) is absolutely continuous, then \(F'\) exists a.e., \(F'\in L^1\), and:

F(x)-F(a)=\int_a^x F'(y)\,dy, \quad a\le x\le b

Conclusion

Differentiating an integral: For \(f\in L^1_{\text{loc}}\), \(F'(x)=f(x)\) almost everywhere via the Lebesgue Differentiation Theorem.
Integrating a derivative: If \(F\) is absolutely continuous, the fundamental relationship is preserved exactly.
Bounded variation: Guarantees a.e. differentiability, but only provides an inequality (\(\le\)) rather than full equality.

← Real Analysis Home Knowledge Base