Is a function $g$ equal to a Lebesgue integrable function $f$ except at points where $f(x)= \pm \infty$ measurable?

Question

I've recently seen a proof of the dominated convergence theorem that states that WLOG we can assume that an integrable function is finite everywhere. I know that an integrable function must be finite $\mu$-almost everywhere and that if $f,g: \Omega \to \overline{\mathbb{R}}$ are measurable and equal $\mu$-almost everywhere, then their integrals on a measurable set $E$ are equal. So I am trying to prove that we can simply change a function $f$ at points $x: f(x)= \pm \infty$ and obtain a measurable function $g$ that is finite everywhere.

More precisely:

Let $(\Omega, \Sigma, \mu)$ be a measure space and let $f: \Omega \to \overline{\mathbb{R}}$ be measurable where we use the Borel $\sigma$-Algebra for $\overline{\mathbb{R}}$, i.e.

$\mathcal{B}(\overline{\mathbb{R}}) = \{B \cup A : B \in \mathcal{B}(\mathbb{R}), A \subset \{-\infty,\infty\}\}$. Note that $\mathcal{B}(\overline{\mathbb{R}})$ is generated by the set $\{(a,\infty]: a \in \mathbb{R}\}$.

Now define a function $g: \Omega \to \overline{\mathbb{R}}$ as follow:

Let $g(x)=f(x)$ for $x$ with $f(x) \in \mathbb{R}$ and $g(x)=0$ for $x$ with $f(x)=\pm \infty$.

Now is $g$ measurable?

My idea is to look at the set $(a, \infty]$ for $x \in \mathbb{R}$ and show that its inverse image

$g^{-1}((a, \infty]) = \{x \in \Omega: g(x)>a\}$

is measurable.

First assume that $a<0$, then

$\{x \in \Omega: g(x)>a\} = \{x \in \Omega: a<g(x)<0\} \cup \{x \in \Omega: g(x)=0\} \cup \{x \in \Omega: 0<g(x)<\infty\} \cup \{x \in \Omega: g(x)= \infty\}$

Also

$\{x \in \Omega: a<g(x)<0\}=\{x \in \Omega: a<f(x)<0\}$,

$\{x \in \Omega: 0<g(x)<\infty\}=\{x \in \Omega: 0<f(x)<\infty\}$,

$\{x \in \Omega: g(x)=0\} = \{x \in \Omega: f(x)=0\} \cup \{x \in \Omega: f(x)=\pm \infty\}$ and

$\{x \in \Omega: g(x)= \infty\} = \emptyset$.

A similar argument holds for $a \geq 0$.

Since $f$ is measurable, all these sets are measurable, and hence $g$ is also measurable.

Is this fine or did I make a mistake?

Answer 1

Showing that $\{x:g(x)>a\}$ are measurable $\forall a$ is correct in principle but not very informative in this context. Instead, consider writing $g(x)=f(x)\mathbf{1}_{\{x:|f(x)|<\infty\}}(x)$. This makes it clear why they wrote 'wlog' (note $\Omega=\{x:|f(x)|<\infty\}\cup\{x:|f(x)|=\infty\}$, and the two sets are disjoint). Since both $f$ and the indicator are measurable, then $g$ is measurable $(*)$.

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$(*)$. To see that the product of measurable functions is measurable: let $u,v$ be measurable, then we can write $uv=((u+v)^2-u^2-v^2)/2$. Since sums of measurable functions are measurable and $u^2,v^2$ are measurable because $y \mapsto y^2$ is continuous (thus measurable), then $uv$ is measurable.