Showing the product of measurable functions is measurable directly from the definition

Question

For simplicity, I assume that $f$ and $g$ are real-valued, Lebesgue measurable functions.

In this question the following is asked:

Prove that if $f$ and $g$ are measurable then $fg$ is also measurable (express the product using sums and powers of functions).

It is not difficult to do this using the fact that $(f+g)^2=f^2+g^2+2fg$, together with (1) $f+g$ is measurable whenever $f$ and $g$ are measurable, and (2) the composition of the continuous function $x\mapsto x^2$ with a measurable function is measurable.

Question. Is it possible to show that the product $fg$ is measurable without knowing these two facts, i.e., directly from the definition?

Edit: I am using the following definition: a function $f:E\to \mathbb R$ on a measurable domain $E\subseteq \mathbb R$ is said to be measurable provided that for every $a \in \mathbb R$, each preimage $f^{-1}((a, \infty))$ is measurable. (or equivalently, each preimage of intervals of the form $[a, \infty), (-\infty, a),$ or $(-\infty, a]$).

Answer 1

Lets consider the $\lambda$-measurable functions $f,g: E \to \overline{\mathbb R}$. Then we must take some care in defining the sum and product of such functions. If $f$ and $g$ are measurable functions, then $f + g$ is undefined at points where it would be of the form $\infty - \infty$.

Let $f$ and $g$ be measurable extended real-valued functions defined on $E$ and $\alpha$ a fixed number. Define $f+g$ to be $\alpha$ whenever it is of the form $\infty - \infty$ or $-\infty + \infty$.

$$ E_1 = \{f+g = \infty\} = \{f \in \mathbb{R}, g = \infty\} \cup \{f = g = \infty\} \cup \{f = \infty, g \in \mathbb{R}\}, $$

which is measurable.

$$ E_2 = \{f + g = -\infty\} = \{f \in \mathbb{R}, g = -\infty\} \cup \{f = g = -\infty\} \cup \{f = -\infty, g \in \mathbb{R}\}, $$

which is measurable.

Let $h = (f+g)|_{E \setminus (E_1 \cup E_2)}$ and let $\beta \in \mathbb{R}$.

If $\beta \geq \alpha$, then

$$ \{x : h(x) > \beta\} = \{x : f|_{E \setminus \{f = \pm \infty\}}(x) + g|_{E \setminus \{g = \pm \infty\}}(x) > \beta\}, $$

which is measurable.

If $\beta < \alpha$, then

$$ \{x : h(x) > \beta\} = \{f = \infty, g = -\infty\} \cup \{f = -\infty, g = \infty\} \cup \{x : f|_{E \setminus \{f = \pm \infty\}}(x) + g|_{E \setminus \{g = \pm \infty\}}(x) > \beta\}, $$

which is measurable. Hence $f + g$ is measurable.

Now we consider the product $fg$, let

$$ E_1 = \{fg = \infty\} = \{f = \infty, g > 0\} \cup \{f = -\infty, g < 0\} \cup \{f > 0, g = \infty\} \cup \{f < 0, g = -\infty\}, $$

which is measurable.

$$ E_2 = \{fg = -\infty\} = \{f = \infty, g < 0\} \cup \{f = -\infty, g > 0\} \cup \{f > 0, g = -\infty\} \cup \{f < 0, g = \infty\}, $$

which is measurable.

Let $h = fg|_{E \setminus (E_1 \cup E_2)}$ and let $\alpha \in \mathbb{R}$. If $\alpha \geq 0$, then

$$ \{x : h(x) > \alpha\} = \{x : f|_{E \setminus \{x : f(x) = \pm \infty\}}(x) \cdot g|_{E \setminus \{x : g(x) = \pm \infty\}}(x) > \alpha\}, $$

which is measurable.

If $\alpha < 0$, then

$$ \{x : h(x) > \alpha\} = \{x : f(x) = 0\} \cup \{x : g(x) = 0\} \cup \{x : f|_{E \setminus \{f = \pm \infty\}}(x) \cdot g|_{E \setminus \{g = \pm \infty\}}(x) > \alpha\}, $$

which is measurable. Hence $fg$ is measurable.