Posted by Jason Polak on 20. August 2011 · 2 comments · Categories: measure-theory

For my qualifying exam next week, I made a few notes on the fundamental theorem of calculus in the Lebesgue setting and I’ve decided to post them in case they might be of use to someone else. I shall sketch the proof and try to explain the main points, aiming for a broad overview. The interested reader should consult Chapter 7 of Rudin’s Real and Complex Analysis for full proofs.

In the Riemann setting, for a continuous Riemann-integrable function $ f:[a,b]\to\mathbb{R}$ differentiable on $ (a,b)$, one form of the fundamental theorem is that

$ \int_a^x f'(t)dt = f(x) – f(a)$.

Here the integral is the Riemann integral. Now, one can make various technical modifications to the hypotheses of the usual analysis ilk, such as allowing $ f$ to be differentiable on $ (a,b)$ except for some finite set, but the above statement seems to capture the essence of the fundamental theorem. Now, the proof of the fundamental theorem is short and sweet: use the definition of the Riemann integral as a limit of Riemann sums and apply the mean value theorem.

We would like a similar statement to hold for the Lebesgue measure $ \mu$ on the real line.

In this case, we would like to find hypotheses that guarantee

$ \int_a^x f'(t)d\mu(t) = f(x) – f(a)$.

Where $ f:[a,b]\to\mathbb{R}$ is a function of some sort, and $ f’$ is the derivative of $ f$. For something like this to make sense, let us start by assuming that $ f’$ actually exists almost everywhere, so at least our formula makes sense. Then $ f’:[a,b]\to\mathbb{R}$ can be any function equal to the derivative of $ f$ when it exists, which is fine for the integral.

A Necessary Condition

The logical thing to do now is assume the formula, and see what kinds of interesting consequences we can squeeze out of it. For any measurable $ E\subseteq [a,b]$, we define

$ \lambda(E) = \int_E f’d\mu.$

If we assume $ f’$ to be positive, then (exercise) $ \lambda$ is a measure on $ [a,b]$ and it is absolutely continuous with respect to $ \mu$. This implies that for any $ \epsilon > 0$ there is a $ \delta > 0$ such that $ \lambda(E) < \epsilon$ for all $ E\subseteq [a,b]$ with $ \mu(E) < \delta$. Now, in particular this holds for $ E = \sqcup^n (a_i,b_i)$, a disjoint union of intervals in $ [a,b]$. If $ \mu(E) < \delta$; that is, if $ \sum (b_i - a_i) < \delta$, then $ \epsilon \geq \lambda(E) = \sum_{j=1}^n\int_{a_i}^{b_i} f'd\mu = \sum_{i=1}^n |f(b_i) - f(a_i)|.$ Hence we can conclude that $ f$ satisfies the following property: for every $ \epsilon > 0$, there is a $ \delta > 0$ such that if $ E = \sqcup (a_i,b_i)$ is a disjoint union of open intervals with total length less than $ \delta$, then $ \sum |f(b_i) – f(a_i)| < \epsilon$. Recall that we have assumed that $ f'\geq 0$, and so the differences $ f(b_i) - f(a_i)$ resulting from the fundamental theorem are nonnegative, and so placing the absolute values around each term did not change the sum. We say that $ f$ is absolutely continuous whenever it has this property. Now, being absolutely continuous is a fairly strong property. In particular, any function that is absolutely continuous is also uniformly continuous, although a continuous function on a closed interval is uniformly continuous anyway.

Absolutely Continuous Functions

The first result in our journey towards the fundamental theorem is the following theorem, which hints that we have stumbled upon an important property:

Theorem 1. If $ f:[a,b]\to\mathbb{R}$ is continuous and increasing, then the following statements are equivalent:

  1. $ f$ is absolutely continous,
  2. $ f$ maps null sets to null sets, and
  3. $ f’$ exists almost everywhere, and $ f(x) – f(a) = \int_a^x f’d\mu$.

We have already observed that $ 3\Rightarrow 1$. Let us now sketch the proof of the remainder of the theorem. For $ 1\Rightarrow 2$, let $ E$ be null. Now, by the outer regularity of the Lebesgue measure, for every $ \epsilon > 0$ there is an open set $ V$ that contains $ E$ and such that $ \mu(V) < \delta$. Now, write $ V$ as a disjoint union of open intervals, apply the absolute continuity of $ f$ to $ f(V)$ and use that $ f$ is increasing. For $ 2\Rightarrow 3$, suppose $ f$ maps null sets to null sets. Define $ g(x) = f(x) +x$ and define $ \lambda(E) = \mu(g(E))$. Then $ \lambda$ is a measure and is represented by some $ h\in L^1(\mu)$. Use this to show that $ f(x) - f(a) = \int_a^x (h-1)d\mu$ and apply the theorem of Lebesgue points. I think the most interesting part of this theorem is that it says almost everywhere, a continuous function is smooth if and only if it maps null sets to null sets. This theorem is our prototype: it is the fundamental theorem, but with the nagging hypothesis that $ f$ be increasing. The usual analysis trick of decomposing a more general function into a sum of simpler functions will work here, although so we will have to work to get our decomposition.

The Decomposition

If we could decompose an absolutely continuous function into a difference of two increasing absolutely continuous functions then we would have our fundamental theorem with the only hypothesis being absolute continuity, which is necessary anyways!

The trick of the decomposition lies in the total variation of a function. If $ f:[a,b]\to\mathbb{R}$ is absolutely continuous, the total variation function $ [a,b]\to\mathbb{R}$ is defined by

$ F(x) = \vee_a^xf = \sup_P \sum_{i=1}^N |f(t_i) – f(t_{i-1})|$,

where the supremum is taken over all partitions $ P$ of $ [a,b]$. The total variation of a function is in general defined for functions of bounded variation, which are defined as functions for which the above supremum is finite. The reason why we define such a function is because of the next theorem.

Theorem 2. If $ f:[a,b]\to\mathbb{R}$ is absolutely continuous and $ F$ its total variation, then $ F$, $ F + f$, and $ F-f$ are absolutely continuous and increasing.

The proof of Theorem 2 is mostly definition pushing sprinkled with a healthy dose of epsilons. The reader should try and prove immediately that $ F +f$ is increasing, as it follows immediately from the definition of the total variation, but the absolute continuity of $ F$ itself requires more work.

By applying Theorem 1 to $ f_1 = F + f$ and $ f_2 = F – f$ and noting that $ f = \tfrac{1}{2}(f_1 – f_2)$ we get the fundamental theorem.

Conclusion

The Lebesgue version is not much harder than the Riemann version. Additionally, we saw that the fundamental theorem formula implies that $ f$ be absolutely continuous, and that this is also sufficient for the fundamental theorem to hold in the Lebesgue sense. Yet, there are absolutely continuous functions that are not integrable in the Riemann sense and so with regard to the Riemann integral, one always has to worry about Riemann integrability. This is certainly yet another facet that exemplifies the elegance of abstract integration theory and the Lebesgue integral.

2 Comments

  1. Michael Snarski

    Nice exposition, clear writing and not condescending. Hope those quals went well.

    A word on solutions to assignments — it’s always a good idea to include the exact statement of the question; that way, the document is self-contained and its worth doubles (it’s always fun to figure out what the question is based on the proof, but not always efficient :)).

  2. Thanks for the comment. You’re right, I’m terrible at including the question, but if the file survives but the question sheet dies it won’t be much use. I’ll definitely do this from now on, however. Good luck with your exams.

    The quals by the way were good. I think it’s pretty rare for anyone to find difficulty with them, no matter where you go. There was a small earthquake when I did mine though!

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