# Gobi

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 1d comment space of riemann integrable functions not complete@Tim Then I should find other examples. But then the above example is just saying that there is a (cauchy) sequence in $\mathcal{R}^1$ such that the pointwise limit is not in $\mathcal{R}^1$? May5 comment Is it necessary to consider inversion $\left(z\mapsto\dfrac{1}{z}\right)$ on the extended complex plane@EricStucky I think that the answer depends on what is the definition of a straight line is. So if the definition does not need to contain the infinity point, you don't need to consider the extended complex plane. Apr23 comment integral on $[a,x]$ is zero for all x implies $f=0$ a.e.@jun Oh I couldn't find it. I should take a look at it. Apr23 comment integral on $[a,x]$ is zero for all x implies $f=0$ a.e.@Easy yes, I'll fix it. Apr22 comment A function with countable discontinuities is Borel measurable.Actually I didn't proved the last statement, so there can be some errors. Apr22 comment A function with countable discontinuities is Borel measurable.@DaveL.Renfro The solution start in there. $f$ is then riemann integrable, so I can define the partition to be increasing set of discontinuous points $P_k$ and lower and upper (Borel-)simple functions $g_{P_k},G_{P_k}$ and the limits of them $g,G$. And use the fact that for $x\in \cup P_k$, $f$ is continuous at $x$ iff $g(x)=G(x)$. Apr22 comment A function with countable discontinuities is Borel measurable.The generalization should be that if all subsets of {discontinuies} are measurable. Apr22 comment A function with countable discontinuities is Borel measurable.Or maybe the conclusion is that if $f$ is satisfying such conditions, then $E$ should be measurable? Apr22 comment A function with countable discontinuities is Borel measurable.I have a question. Your answer can be generallized to for any $f:E \to \mathbb{R}$ with the set of discontinuities measurable, then $f$ is measurable. But suppose that $E$ is not measurable. Your proof does not use any condition about $E$. Then \$E=\cup \{ f