Intuition for Lebesgue integration

Question

I have started doing Lebesgue integration and I just want to clarify one thing to start with.

Very loosely speaking, with Riemann integration we partition the domain into $n$ intervals, and then we calculate the area of the $n$ rectangles formed by a base of width $n$ and a height of where each rectangle 'hits' the function. Summing these rectangles gives us an approximation of the area of the graph under the function. Then as we let $n \to \infty$ this approximation increase in accuracy and equals the function when we take the limit.

Now with Lebesgue integration do we follow the same process of partitioning (the range this time) into $n$ intervals and then letting $n \to \infty$ giving us smaller and smaller intervals which implies the approximation to the function keeps increasing? Leaving aside the concept of having sets that are measurable on the domain...I am simply wondering is process of considering intervals of decreasing size the same as with Riemann integration?

Answer 1

Suppose that $f$ is a positive bounded function defined on an interval $[a,b]$ with $0 < f \le M$.

If you partition the domain using $a = t_0 < t_1 < t_2 < \cdots < t_n = b$ you select points $x_k \in [t_{k-1},t_k]$ and form the Riemann sum $\displaystyle \sum_{k=1}^n f(x_k) \ell([t_{k-1},t_k])$ where $\ell$ is the length of the interval $[t_{k-1},t_k]$.

If instead you partition the range using $0 = M_0 < M_1 < M_2 < \cdots < M_n = M$, you can form an analogous "Lebesgue sum" by defining $E_k = \{x \in [a,b] : M_{k-1} < x \le M_k\}$. Select points $x_k \in E_k$ and use the sum $\displaystyle \sum_{k=1}^n f(x_k) \ell(E_k)$.

There are two things going on that are much different than in the Riemann sum. First, the sets $E_k$ aren't necessarily intervals so you have to be careful in what is meant by $\ell(E_k)$. This is where Lebesgue measure is needed. Second, even if the $E_k$ are intervals, their lengths don't necessarily shrink to zero as the mesh of the partition of $[0,M]$ decreases to zero. So, the process of considering intervals of decreasing size is not the same as with Riemann integration.

Answer 2

The essence can be better understood in a two-dimensional setting. The Riemann integral of a function $(x,y)\mapsto f(x,y)$ over the square $Q:=[-1,1]^2$ involves cutting up the square into small rectangles $Q_{ij}:=[x_{i-1},x_i]\times[y_{j-1},y_j]$ and then arguing about "Riemann sums" of the form $$\sum_{i, j} f(\xi_i,\eta_j)\mu(Q_{ij})\ ,\tag{1}$$ where $\mu(Q_{ij})=(x_i-x_{i-1})(y_j-y_{j-1})$ is the elementary euclidean area of $Q_{ij}$. This is simple and intuitive, and you get the linearity of the integral for free. But it is somewhat rigid.

Contrasting this, Lebesgue integration involves cutting up the square in subsets defined by the level curves of $f$, like so:

The integral then is the limit $N\to \infty$ of sums of the form $$\sum_{k=-\infty}^\infty {k\over N}\mu(S_{N,k})\ ,\tag{2}$$ where $S_{N,k}$ is the zone in the figure where $${k\over N}\leq f(x,y)<{k+1\over N}\ .$$ It should be intuitively clear that the sums $(2)$ are in the same way an approximation to the volume of the cake defined by $f$ over $Q$ as are the sums $(1)$. But the approach $(2)$ is much more flexible and allows for more powerful "limit theorems". On the other hand such "simple" things as linearity now require proof.