Lecture 10: Measurable functions I

31/10/24

Reference: Lecture notes sections 5

Recall from the first lecture that the basic example of a function that is not Riemann integrable is the Dirichlet function $f \colon [0,1] \to \R$ given by $f(x) = 1$ if $x\in\mathbb{Q}$ and $f(x) = 0$ if $x \notin \mathbb{Q}$. The problem with this function is that no matter how finely we divide $[0,1]$ into smaller intervals, the lower Riemann sum is always $0$ while the upper Riemann sum is always $1$.

On the other hand, we would want to say that the integral of $f$ is $0$. One reason $f$ is the pointwise limit of a sequence of functions whose integral is $0$, namely $f_n \colon [0,1] \to \R$ which take values $1$ on the first $n$ rational numbers, with respect to some enumeration, and $0$ otherwise. The Dirichlet function shows that the property of being Riemann integrable is not preserved under limits.

The second reason we would want $\int_{[0,1]} f = 0$ is the following. For every set $A \subset \R$ denote by $\chi_A \colon \R \to \R$ its characteristic function defined by $\chi_A(x) = 1$ if $x \in A$ and $\chi_A(x)=0$ if $x \notin A$. If $A$ is an interval, then trivially

$\int_\R \chi_A = l(A)$

is the length of $A$. By now, we know that it is useful to generalize length to the Lebesgue measure $\lambda$. So it would be reasonable to have a notion of the integral such that for every $A$ which is Lebesgue measurable, we have

$\int_\R \chi_A = \lambda(A)$

In particular, since the Dirichlet function is $\chi_\mathbb{Q}$ we would want

$\int_\R \chi_{\mathbb{Q}}= \lambda(\mathbb{Q})=0$

since $\mathbb{Q}$ is a countable set. In this lecture we will introduce the Lebesgue integral which indeed satisfies this property. The basic difference between the Riemann integral and the Lebesgue integral is: instead of dividing the domain of a function, divide the range of the function.This intuitive interpretation is explained well in the first two sections of this article:

https://en.wikipedia.org/wiki/Lebesgue_integral

Definition. Let $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ be measurable spaces. A function $f \colon X \to Y$ is measurable with respect to $(\mathcal{A}, \mathcal{B})$ (or $(\mathcal{A},\mathcal{B})$-measurable) if for every $B \in \mathcal{B}$ it satisfies $f^{-1}(B)\in\mathcal{A}.$ In that case, we write

$f \colon (X,\mathcal{A}) \to (Y,\mathcal{B})$

and if it is clear from the context what $\mathcal{A},\mathcal{B}$ we mean we call it simply measurable.

Remark. Compare with the definition of a continuous map between two topological spaces: if $(X,\mathcal{T}_X)$ and $(X,\mathcal{T}_Y)$ are topological spaces, then $f \colon X \to Y$ is continuous if the preimage of every open set is open, that is: $f^{-1}(U) \in \mathcal{T}_X$ for every $U \in \mathcal{T}_Y$.