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.
 A: Let $f,g:E\subset \mathbb R^n\to\overline{\mathbb R}$ be two measurable functions, then $\tilde E=\{-\infty<f,g<+\infty\}$ 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 $\forall\beta\in\mathbb R$.
A: 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.
A: 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.
A: 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.
