## Borel Measurable

This is a continuation of the study of the book The Elements of Integration and Lebesgue Measure by Bartle, listing a few examples of functions that are measurable. Bartle is a very good author, he tries his very best to make this difficult subject accessible to undergraduates.

Example:

If X is the set R of real numbers, and X is the Borel algebra B, then any monotone function is Borel measurable.

Proof:

Suppose that f is monotone increasing, i.e. $x\leq x'$ implies $f(x)\leq f(x')$.

Then, $\{x\in\mathbb{R}:f(x)>\alpha\}$ consists of a half-line which is either of the form $\{x\in\mathbb{R}:x>a\}$ or the form $\{x\in\mathbb{R}:x\geq a\}$. (We will show later that both cases can occur.) Thus,  the set will belong to the Borel algebra B which is the $\sigma$-algebra generated by all open intervals (a,b) in R.

Both cases can indeed occur. For example, if f(x)=x, then the set will be of the form $\{x\in\mathbb{R}:x>a\}$. More interestingly, if the set is the step function $f(x)=\begin{cases}-1, &\text{if }x<0\\1, &\text{if }x\geq 0\end{cases}$, then when $\alpha=0$, the set will be $\{x\in\mathbb{R}:x\geq 0\}$.

Lemma: An extended real-valued function f is measurable if and only if the sets $A=\{x\in X:f(x)=+\infty\}$, $B=\{x\in X:f(x)=-\infty\}$ belong to X and the real-valued function $f_1$ defined by $f_1 (x)= \begin{cases} f(x), &\text{if }x\notin A\cup B,\\ 0, &\text{if }x\in A\cup B,\end{cases}$ is measurable.

This lemma is often useful when dealing with extended real-valued functions.

Proof: If f is in M(X,X), it is proven earlier in the book by Bartle that A and B belong to X. Let $\alpha\in\mathbb{R}$ and $\alpha\geq 0$, then we have that $\{ x\in X:f_1 (x)>\alpha\}=\{ x\in X:f(x)>\alpha\}\setminus A$ which is in X since it is the complement of the union of A and $X\setminus \{x\in X:f(x)>\alpha\}$.

If $\alpha<0$, then $\{ x\in X:f_1 (x)>\alpha \}=\{ x\in X:f(x)>\alpha \}\cup B$, which is a union of two sets in X and hence also in X.

Hence, $f_1$ is measurable.

Conversely, if $A, B\in \mathbf{X}$ and $f_1$ is measurable, then $\{x\in X:f(x)>\alpha\}=\{ x\in X: f_1 (x) >\alpha \}\cup A$ when $\alpha \geq 0$, and $\{x\in X:f(x)>\alpha\}=\{x \in X:f_1 (x)>\alpha\}\setminus B$ when $\alpha <0$, due to a similar reason as above. Therefore f is measurable!