Lebesgue Integral definition In $\mathbb{R}^n$

Question

I am studying Lebesgue Integration in $\mathbb{R}^n$, reading chapters from different books, but still not sure what is Lebesgue integral! Is there an explicit definition for Lebesgue integral, or an idea to understand it?

Answer 1

The Lebesgue integral is a way to integrate with respect to measures. Measures are in turn an abstract way to discuss the idea of "size" of a set for the purpose of integration theory. I feel like it's probably best to motivate through the following example.

Consider the function $ f(x) = \begin{cases} 1 & \text{if}\ 0 \le x \le 1 \\ -1 & \text{if } 1 < x \le 2 \\ 1 & \text{if } 2 < x \le 4 \end{cases} $

I'd like to evaluate the integral

$ \int_0^4 f(x) dx$

There are two ways to integrate this. The first is to note that

$\int_0^4 f(x) dx = \int_0^1 f(x) dx + \int_1^2 f(x) dx + \int_2^4 f(x) dx$

$= 1 \cdot\text{length}([0,1]) + (-1)\cdot \text{length}([1,2]) + 1 \cdot \text{length}([2,4]) \\ = 1 - 1 + 2 = 2$.

This corresponded to decomposing the domain of $f$ into intervals and taking a value for $f$ over each interval, together with a notion of what the length of an interval ought to be. This is essentially the guiding philosophy in the theory of Riemann integration.

Here is another way to do the same integral.

$\int_0^4 f(x) dx \\ = \int_{[0,1] \cup [2,4]}f(x) dx + \int_1^2 f(x) dx \\ = 1 \cdot \text{length}([0,1] \cup [2,4]) + (-1) \cdot \text{length}([1,2]) \\ = 3 - 1 = 2$.

This method of integration was done by breaking the range of $f$ into its two components ($1$ and $-1$) and then evaluating the length of the region that $f$ takes those values and adding them together. This is the theory of Lebesgue integration. Note that it immediately requires more material than Riemann integration will; to perform a Riemann integral, I need to know what the length of an interval will mean. To do even this simple integral using the Lebesgue philosophy, I need to know what the length of the disjoint union of two intervals is.

We were saved above because we do really know what the length of an interval should mean (we should have that $\text{length}([a,b]) = b-a$), and we know that length should be additive under disjoint unions. Let's try and think about what we'd really like out of a theory for measuring sets:

From the above, we know that we want $\text{length}([a,b]) = b - a$. We'd also really like the $\text{length}( \{a\}) = 0$, for any point; a point should have no length. We'd like that our length is invariant under translations - $\text{length}(A + x) = \text{length}(A)$. We'd like that we can always tell the length of any set. And we'd also like additivity in some sense. That is, we'd like to have the $\text{length}(\bigcup_{\alpha} E_{\alpha}) = \sum_{\alpha} \text{length}(E_{\alpha})$. We should be careful about how additive things should be. Since we've decided that a point has length $0$, we cannot have arbitrary additivty, or everything must have $0$ length. But this is analysis and we'd like to discuss limits, so we'll at least require countable additivity. That is, $\text{length}(\bigcup_n E_n) = \sum_n \text{length}(E_n)$.

It turns out that there is no theory of length on $\mathbb{R}^n$ that will give us what we've asked for. We must give something up. It turns out we'll give up the requirement that every set has a length. The resulting theory will lead to the theory of Lebesgue measure, which we can use to define a Lebesgue integral.

The fact that we must restrict ourselves to a convenient subclass of sets to measure is why we need a "measure theory" in the first place, and it's why we study it (at least initially; maybe some people find it fascinating...maybe you do!).

I hope that answers your question and gives you a bit of motivation!

Edit: I suppose it doesn't quite answer your question what the Lebesgue integral is. To do that, you will of course need to read your books and study the definitions therein. My intention here was to try to motivate why one would study abstract measures and an abstract integral in the first place.

Answer 2

There are many equivalent ways of defining the Lebesgue integral. Basically they are all variants of this concept: Given a set $V$ in $\mathbb R^n$, you can cover it with a countable number of n-cubes of various sizes. We know what the n-volume of n-cubes should be (side-length raised to the nth power), so define the size of the cover to the sum of the n-volumes of the cubes. (This may be $\infty$.) Obviously, the n-volume of $V$ should be less than the n-volume of any cover of it, so take the infimum of the sizes of all covers of $V$. This is Lebesgue outer-measure $\mu$. On common sets, it is well-behaved, but some really esoteric ones, it behaves badly, so we toss in some technical conditions to restrict the sets $V$ to those where it is well-behaved (the Lebesgue $\sigma$-algebra), and carefully avoid the rest.

Then for $f: \mathbb R^n \to \mathbb R$, $\int f d\mu = \mu(\{ (x,y) \in \mathbb R^n \times \mathbb R | 0 \le y \le f(x)\}) - \mu(\{ (x,y) \in \mathbb R^n \times \mathbb R | f(x) \le y \le 0\})$