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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.

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

4 Answers 4

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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.

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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 $\forall\beta\in\mathbb R$.

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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.

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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.

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