Importance of Measure Theory

Question

I know that the whole concept of Measure was introduced to develop Lebesgue integration. But I just wondered Even if we had Riemann integration already why it was necessary .

Thanx and regards in advance

Answer 1

The concept of a measure goes far beyond integration on R^n. For example, it underlies the probabilists notion of random. This includes: brownian motion and stock markets, point processes (modelling the locations of rain drops), infinite sequences of coin flips, statistical tests, etc.

It's also important in geometry, where having a notion that generalizes surface area is very useful. (E.g. fix a box in 3 space. What is a random line through this box, and what are it's properties? This problem is solved by first introducing a measure on the space of lines. This apparently turns out to be a useful notion in stereology.)

Having a general theory about measures is therefore very useful, since you don't have to start from scratch with each of these examples.

Even on $R^n$ the notion is very useful - the vector space of Lebesgue integrable functions has some very convenient properties (such as completeness), which makes certain constructions easier as it makes it simpler to check that a sequence has a limit.

Also, it's pretty hard to build functions which you can't plug into the machinery of the Lebesgue measure (which some manipulations maybe), but the same is not true of the Riemann integral. This means that there are simpler hypothesis to check in theorems. Along these lines, there are some very strong theorems with relatively weak assumptions that hold for the Lebesgue integral, but not the Riemann integral (unless you impose additional assumptions) - such as the Lebesgue dominated convergence theorem.

Answer 2

An important difference between the Riemann integral and Lebesgue integral is with regard to pointwise limits.

Consider the sequence of functions $f_n:[0,1] \to \mathbb{R}$ defined by

$$f_n(x) = \begin{cases} 1 &\text{if $x\in\mathbb{Q}$ with $x=\frac{p}{q}$, $q \leq n$}\\0 & \text{otherwise}\end{cases}$$ i.e. $f_n(x)$ is $1$ only if $x$ is a rational number with denominator at most $q$. Now, each $f_n$ is $0$ at all but finitely many points, so they are integrable and $\int_0^1 f_n \, dx = 0$.

But, the pointwise limit $f(x) :=\lim_{n\to\infty} f_n(x)$ is given by $$f(x) = \begin{cases} 1 &\text{if $x\in\mathbb{Q}$}\\0 & \text{otherwise}\end{cases}$$ the indicator function of the rationals. However $f(x)$ is not Riemann integrable, since for any partition, on every interval in the partition, the supremum of $f$ is $1$ and the infimum is $0$, so the lower and upper integrals always differ by $1$.

The Lebesgue integral restores this nice property (partially) with the Beppo-Levi monotone convergence theorem and the Lebesgue dominated convergence theorem which say that if $f$ is the pointwise limit of functions $f_n$ and either $f_n$ is a pointwise increasing sequence (as above) or $f_n$ are all bounded by some integrable function $g$ (again, as above with $g=1$) then

$$\lim_{n\to\infty}\int f_n(x)\, dx=\int\lim_{n\to\infty}f_n(x) \, dx.$$ More generally, we also have theorems allowing swapping of differentation and integration and swapping the order of integration.

In addition, the Lebesgue integral actually generalises the notion of a sum as well, and so many theorems seemingly about integration in fact apply to summation also, using the counting measure.

Answer 3

Measure theory arises naturally when you attempt to generalize the notion of the size of a set, such that it has the expected features: it's additive, non-negative and the size of $\emptyset$ is $0$. If you try to construct different measures on $\mathbb{R}^n$, you quickly find out there is essentially only one measure that generalizes the lengths of intervals, and that is also translation-invariant; namely, the Lebesgue measure $\lambda$.

Remember, when you calculate the Riemann sums of a function $f : [a,b] \rightarrow \mathbb{R}$ for the integral, you make use of the lengths of subintervals:

$$\sum_{i=o}^{n-1}f(\xi_i)(x_{i+1} - x_i)$$

for some interval division and set intermediate points. How would you do this if $f$ were defined on an arbitrary space $X$ that has nothing to do with the real line?

On differentiable manifolds, except for the trivial cases, it is also easy to show that integration of functions is ill-defined, even though the manifolds locally look like $\mathbb{R}^n$. You need additional structure. If the manifold is orientable, you can achieve this by prescribing a volume form.