Let X be a set and $\sum$ a $\sigma$ algebra of subsets of X. Let f and g be real valued functions defined on domains :
dom $f$ and dom $g$ $\subseteq X$.
If $f$ and $g$ are measurable , so is $f+g$ where $(f+g)(x)=f(x)+g(x)$ for $x \in$dom$f \cap $ dom$g$
What does dom mean in this context?
What are dom $f$ and dom $g$? And why are they so significant?
$\endgroup$3 Answers
$\begingroup$dom is short for domain (of a function).
$\endgroup$ $\begingroup$The statement
$f(x)=\frac1x$ is differentiable with $f'(x)=-\frac1{x^2}$
is not precise, it has a problem. But the statement
$f(x)=\frac1x$ is differentiable on its domain, $dom{f}=\mathbb R\setminus \{0\}$ with $f'(x)=-\frac1{x^2}$
is correct. Similarly (you can construct many, many examples) in your case consider the statements:
Let $f(x)=\frac1x$ and $g(x)=x$, so $f(x)+g(x)=x+\frac1x$
So, $g(0)=0$ but what is $(f+g)(0)$? It is not defined. So, improve (considerably) the previous statement as
Let $f(x)=\frac1x$ for any $x$ except $x=0$ (domain of $f$ is $\mathbb R\setminus\{0\}$ and $g(x)=x$ for any $x$ (domain of $g$ is $\mathbb R$), so $(f+g)(x)=x+\frac1x$ for any $x$ except $x=0$.
The main problem is that outside the domain of the function, statements about the function do not make sense.
$\endgroup$ 6 $\begingroup$Dom means domain. A function may not be defined on all of $X$ but only on a subset.
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