Thomas E.
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 Apr22 answered What are possible variations of the definition of $\sigma$-additivity? Apr22 comment Integral of an $L^2$ function Maybe one comment worth mentioning is that you can't use the usual $L^{2}(\mathbb{R})$ Cauchy-Schwarz inequality, since the constant function $1$ is not in $L^{2}(\mathbb{R})$. But instead, for each fixed $x,y\in \mathbb{R}$ you may consider $f$ and the constant function $1$ restricted to $[x,y]$ and the Cauchy-Schwarz related to $L^{2}([x,y])$. But anyways, it doesn't really change anything, just a technical detail. Apr22 comment Is a Metric space $(X,d)$ with $X=\{x\}$ an open set? Where in the definition does it require an $y\in X$ so that $d(x,y)=r$, $r>0$? In this case any open ball centered at $x$ would equal the singleton $\{x\}$. Apr22 revised Basic proof that $T$ is a topology. In the question, there was a mistake in the set-notation: /in instead of /subset Apr22 suggested approved edit on Basic proof that $T$ is a topology. Apr22 comment Showing a function is not uniformly continuous Yeah, you're right. I edited my post now. Apr22 revised Showing a function is not uniformly continuous deleted 34 characters in body Apr22 comment Showing a function is not uniformly continuous Except that is the last $-\frac{1}{\delta^{4}}$ missing from there? Otherwise it looks close to mine. Apr22 revised Showing a function is not uniformly continuous added 20 characters in body Apr22 comment Showing a function is not uniformly continuous I also edited now my calculation which still differs abit from your new one. Could you show the steps of how you got this answer for $|f(x)-f(y)|$? Apr22 revised Showing a function is not uniformly continuous added 1066 characters in body Apr22 revised Showing a function is not uniformly continuous added 132 characters in body Apr22 answered Showing a function is not uniformly continuous Apr22 revised Set Theory - Subset of set added 122 characters in body Apr22 comment Set Theory - Subset of set For $\{\{1\}\}$ to be a subset of the third element $\{1,2\}$, if I understand what you mean, then it would require $\{1\}$ to be an element of $\{1,2\}$, which is not the case. And by the way, even if it would be a subset of the third element, it would not imply that it would be a subset of the whole collection.. Apr22 answered Set Theory - Subset of set Apr21 comment A problem concerning outer measure I posted a counter-example now, see my reply. Apr21 answered A problem concerning outer measure Apr21 comment A problem concerning outer measure This proof is ofcourse valid only for regular outer measures, such as Lebesgue outer measure. But does not hold in the general case for every outer measure $m^{*}$... Apr21 comment A problem concerning outer measure That proof fails in the assumption that for each $k\in\mathbb{N}$ there exists a measurable cover $H_{k}$ of $E_{k}$, i.e. the first verse. This would require $m^{*}$ to be regular. In the Lebesgue case it holds since the Lebesgue outer measure is even Borel regular (by choosing a $G_{\delta}$ cover...). If you read the textbook, it is very explicit that this proof is for Lebesgue outer measure.