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 <n,\\ \emptyset, &\text{if }\alpha\geq n,\\X, &\text{if }\alpha\leq -n \end{cases}

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!

Source: The Elements of Integration and Lebesgue Measure

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