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!

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