Measurability of product fg

In the previous chapters, Bartle showed that that if f is in M(X,X), then the functions $cf, f^2, |f|, f^+, f^-$ are also in M(X,X).

The case of the measurability of the product fg when f, g belong to M(X,X) is a little bit more tricky. If $n\in\mathbb{N}$, let $f_n$ be the “truncation of f” defined by $f_n (x)=\begin{cases}f(x), &\text{if }|f(x)|\leq n, \\ n, &\text{if } f(x)>n,\\ -n, &\text{if }f(x)<-n\end{cases}$

Let $g_m$ be defined similarly. We will work out the proof that $f_n$ and $g_m$ are measurable (Bartle left it as Exercise 2.K).

Proof:

Each $f_n$ is a function on $X$ to $\mathbb{R}$.

$\{x\in X:f_n (x) >\alpha\}=\begin{cases}\{x \in X: f(x)>\alpha\}, &\text{if }-n<\alpha

All of the above sets are in X.

Thus, we may use an earlier Lemma 2.6 to show that the product $f_n g_m$ is measurable.

We also have $f(x)g_m (x)=\lim_n f_n (x)g_m (x)$, and using an earlier corollary that says that if a sequence $(f_n)$ is in M(X,X) converges to f on X, then f is also in M(X,X), we have that $f(x)g_m (x)$ belongs to M(X,X).

Finally, (fg)(x)=f(x)g(x)=$\lim_m f(x)g_m (x)$, and hence fg also belongs to M(X,X).

This is a very powerful result of Lebesgue integration, since we can see that the theory includes extended real-valued functions, and prepares us to integrate functions that can reach infinite values!