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The Fundamental Theorem of Calculus 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 (and for me to prove). 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:

\text{If \(f\) is an integrable function on } [a,b], \quad F(x) = \int_a^x f(y)\,dy,

\text{can we conclude that } F'(x) \text{ exists, and that } F'(x)=f(x) \text{ (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 and the Lebesgue Differentiation Theorem

We define

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

To see if the derivative exists, 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

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

This raises the averaging problem: does the average value over shrinking neighborhoods converge to \(f(x)\)?

Continuous Case

If \(f\) is continuous, then indeed

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

The proof follows from continuity, rewriting the difference as \(\tfrac{1}{m(B)}\int_B(f(y)-f(x))dy\), and applying the triangle inequality.

The General Case

Lebesgue Differentiation 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.

Local Integrability and Density

Local integrability (\(f\in L^1_{\text{loc}}(\mathbb{R}^d)\)) suffices, since behavior at infinity does not affect averages on finite balls. This leads to notions of density and Lebesgue sets.

Lebesgue Density Theorem
For a measurable set \(E\subset\mathbb{R}^d\), almost every point of \(E\) is a point of density of \(E\), and almost every point not in \(E\) is not a point of density.

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 when

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

holds. Not all continuous functions are differentiable everywhere (e.g. the Weierstrass function). We need conditions ensuring differentiability and integrability of \(F'\).

Bounded Variation

Definition (Bounded Variation) \(F\) has bounded variation on \([a,b]\) if \[ \sum_{j=1}^N |F(t_j)-F(t_{j-1})| \le M \] for all partitions \(a=t_0<\cdots<t_N=b\).

Functions of bounded variation are differentiable a.e., but the equality above may fail (e.g. the Cantor function).

Differentiation Existence Theorem
If \(F\) has bounded variation on \([a,b]\), then \(F'(x)\) exists for almost every \(x\).

However, one only obtains

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

Absolute Continuity

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

The Fundamental Theorem of Calculus splits into two directions:

Differentiating an integral: for \(f\in L^1_{\text{loc}}\), \(F'(x)=f(x)\) almost everywhere (Lebesgue Differentiation Theorem).
Integrating a derivative: if \(F\) is absolutely continuous, then \(F(b)-F(a)=\int_a^b F'(x)\,dx\).

Bounded variation guarantees a.e. differentiability but not full equality, while absolute continuity provides the exact condition. This completes the rigorous version of the FTC in the Lebesgue setting.