Inherent Pitfall of Lebesgue Integration?

Question

I am studying Real Analysis with Royden's Book. I noticed that for a function $f$ differentiable almost everywhere on $[a, b]$ and $f'$ integrable over $[a, b],$ it does not imply that $ f(x) = \int_{[a, x]}f'(t)\,dt + f(a).$ Take Cantor function $f$ as a counter example, since $1 = f(1) > \int_{[0, 1]}f'(t)\,dt + f(0) = 0,$ because $f'(x) = 0 $ $ \forall x \in [0,1]$. However, does it mean that Lebesgue integral has some inherent pitfall in the definition of integration? In a sense that it fails to restore the original picture of a function in a set of measure zero after differentiation. Is that any better (or more general) design of integration? so that we can guarantee $ f(x) = \int_{[a, x]}f'(t)\,dt + f(a)$.

Answer 1

That's the distinction between a function being everywhere differentiable and almost everywhere differentiable. You can still recover something from almost everywhere differentiability if you impose the extra condition of absolute continuity. It's a stronger version of continuity that excludes pathologies like the Cantor function. If $f$ is absolutely continuous, then it has a derivative almost everywhere and $ \displaystyle f(x) = f(a) + \int_a^x f^{\prime}(t) \, dt$.

Answer 2

When we get to Lebesgue integration, we typically do so because we have identified pathologies in which the (conventional) Riemann integration fails. It is a natural question to ask whether the Lebesgue integral belongs to a measure theoretic version of the Fundamental Theorem of Calculus. And while at first it may not appear to be a big deal, it doesn't really make sense to attempt to establish the fundamental theorem of calculus using measure-theoretic (Lebesgue) integration and conventional differentiation.

Indeed, the standard way of defining a derivative uses limits of ratios of differences; differences make sense when we have well-defined sets and the notion of measure coincides with our notion of 1-dimensional absolute value. So it instead might make sense to explore not a different kind of integration, but a different kind of differentiation that works in a measure theoretic sense.

This is in fact the essence of the Radon-Nikodym derivative. This derivative uses the fact that given a measurable set $E$ and a measurable function $f$, the mapping $E \mapsto \int_E f\, d\mu$ is a signed measure, and we denote this relationship with the notation $d\nu = fd\mu$; moreover, it is true that given $\sigma$-finite measures $\mu$, $\nu$, with $\nu$ positive and $\nu = 0$ whenever $\mu = 0$, that we can write $$d\nu = \frac{d\nu}{d\mu}d\mu.$$ We call $\frac{d\nu}{d\mu}$ the Radon-Nikodym derivative. Note, this is actually a special case of a slightly more broad statement.

Using a bit more care in restricting our domain, we can actually develop an analog to the Fundamental Theorem of Calculus, known as the Lebesgue Differentiation Theorem, which states that for a locally integrable function $f$, for a.e. $x$, we have $\lim_{r\to 0} \frac{1}{m(E_r)} \int_{E_r} f(y)\, dy = f(x)$ for every $E_r$ that "shrinks nicely" to $x$.

So we can do what you want to do, but we have to use a notion of the derivative that melds with what we do in a measure theoretic sense. You can't play basketball with hockey pucks, but you can sometimes play both in the same arena.

Answer 3

Two things:

1. If $f$ is everywhere differentiable on $[a,b]$ and $f'\in L^1([a,b]),$ then $f(b)-f(a) = \int_a^b f'.$

2. There exists an everywhere differentiable, strictly increasing function $f$ on $[a,b]$ with $f'$ bounded, such that $f'$ is not Riemann integrable on $[a,b].$ The Lebesgue integral doesn't care: $f(b)-f(a) = \int_a^b f'$ still holds by 2.