# Measurable functions $f,g$ are finite a.e. Then, $f+g$ is measurable.

Let $$E\in \mathbb{R}^n$$ be a Lebesgue measurable set.

Let $$f,g:\mathbb{R}^n \to \overline{\mathbb{R}}$$ be Lebesgue measurable functions. Suppose $$f$$ and $$g$$ are finite almost everywhere. Then, prove that $$f+g$$ is a Lebesgue measurable function.

In $f+g$ is measurable no matter how it is defined at points where it has the form $\infty-\infty.$?, use the fact that if $$f$$ is measurable and $$f=g$$ a.e., then $$g$$ is measurable.

But I don't know how I should use the fact.

Thank you for your help. Other way to prove is also welcomed.

• Define $\eta = (f, g)$ and now $f + g = h \circ \eta$. Now we have $(f + g)^{-1} (a, \infty) = \eta^{-1} (h^{-1} (a, \infty))$.
– E.E.
Jun 20, 2021 at 10:07

Suppose $$f,g: \mathbb R^n \to \mathbb R\cup\{\pm\infty\}.$$

Let $$h(x) = \begin{cases} g(x) & \text{if } g(x)\in\mathbb R, \\ 0 & \text{if } g(x)\in\{\pm\infty\}. \end{cases}$$

Now use the fact that if $$g$$ is measurable and $$g=h$$ a.e., then $$h$$ is measurable.

You have $$f+g= f+h$$ a.e. So the problem of showing $$f+g$$ is measurable is reduced to that of showing $$f+h$$ is measurable, and here you have no $$\infty-\infty$$ problem.

Let $$f,g:E\subset \mathbb R^n\to\overline{\mathbb R}$$ be two measurable functions, then $$\tilde E=\{-\infty is measurable.
So for $$\beta\in\mathbb R$$ we have $$\{f+g>\beta\}\cap\tilde E=\{x\in\tilde E: f(x)>\beta-g(x)\}=\bigcup_{r\in\mathbb Q}(\tilde E\cap\{f>r\}\cap\{g>\beta-r\})$$ is measurable $$\forall\beta\in\mathbb R$$.

An alternative way is to take nonnegative simple functions $$\phi_j \nearrow f$$, $$\psi_j \nearrow g$$. Then $$\phi_j + \psi_j \nearrow f + g$$, so $$f + g$$ is a limit of the measurable functions $$\phi_j + \psi_j$$, and is therefore measurable. This also works for proving $$fg$$ is measurable.

The mapping $$\phi:(s,t)\mapsto s+t$$, from $$\Bbb R^2$$ to $$\Bbb R$$ is continuous, hence Borel measurable. On the set $$H:=\{x\in\Bbb R^n: f(x)\in\Bbb R, g(x)\in\Bbb R\}$$ (given the relative Lebesgue measurable sets), the mapping $$\psi:x\mapsto (f(x),g(x))$$, into $$\Bbb R^{2}$$ with the Borel sets, is measurable. Therefore the composition $$x\mapsto f(x)+g(x) = \phi(\psi(x))$$ is Lebesgue measurable on $$H$$. The assertion follows from the completeness of Lebesgue measure because the complement of $$H$$ is null.