# catamite

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210
bio website reddit.com/r/… location meatspace age member for 1 year, 11 months seen 5 hours ago profile views 1,026

when someone smiles at me, all I see is an ape bearing its teethe

# 573 Actions

 May30 comment Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$ @srijan this is essentially what I concluded about the function in my post. But I still don't see how there can be a set of measure zero on which the integral of $f'$ is non-zero. May30 comment Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$ @Shuhao I know it's not of bounded variation, but I can't see how the measure theoretic definition of not being absolutely continuous can possibly be satisfied. May30 revised Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$ deleted 9 characters in body May30 asked Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$ May29 accepted Rigorous proof that $\int_{\Omega}X\;dP=\int_{-\infty}^{\infty}xf(x)\;dx$ May29 comment Rigorous proof that $\int_{\Omega}X\;dP=\int_{-\infty}^{\infty}xf(x)\;dx$ It took me a while to fully digest this but I now see that it is exactly the answer I was looking for, thanks. May29 accepted For $\mu(X)<\infty$ and $f_n<\infty$ a.e. s.t. $\forall M$ $\exists n$ s.t. $\mu(\{f_n>M\})\geq\delta$, then $f_n$ diverges on set of positive measure May29 comment For $\mu(X)<\infty$ and $f_n<\infty$ a.e. s.t. $\forall M$ $\exists n$ s.t. $\mu(\{f_n>M\})\geq\delta$, then $f_n$ diverges on set of positive measure Brilliant, exactly what I was looking for, I knew there had to be some theorem which used the fact that $X$ was finite, and there it was, descending continuity of measure, hiding right under my nose the whole time. Thanks. May28 asked For $\mu(X)<\infty$ and $f_n<\infty$ a.e. s.t. $\forall M$ $\exists n$ s.t. $\mu(\{f_n>M\})\geq\delta$, then $f_n$ diverges on set of positive measure May26 comment Rigorous proof that $\int_{\Omega}X\;dP=\int_{-\infty}^{\infty}xf(x)\;dx$ oh ok I see, thanks May26 comment Rigorous proof that $\int_{\Omega}X\;dP=\int_{-\infty}^{\infty}xf(x)\;dx$ but what does it mean when you're integrating with respect to $t$? I can't figure out how that equals $X$ May26 comment Rigorous proof that $\int_{\Omega}X\;dP=\int_{-\infty}^{\infty}xf(x)\;dx$ could you explain what $\int_0^{\infty}1_{t\leq X}dt$ means? May26 revised Rigorous proof that $\int_{\Omega}X\;dP=\int_{-\infty}^{\infty}xf(x)\;dx$ added 14 characters in body May26 asked Rigorous proof that $\int_{\Omega}X\;dP=\int_{-\infty}^{\infty}xf(x)\;dx$ May25 accepted An alternate proof of Egorov's Theorem May25 comment An alternate proof of Egorov's Theorem you're right Norbert it isn't. May25 comment An alternate proof of Egorov's Theorem Ah yes I see what you mean. Yah I think you're right. So the lesson learned is that I don't want the size of my set to depend on epsilon. May25 revised An alternate proof of Egorov's Theorem added 38 characters in body May25 revised An alternate proof of Egorov's Theorem added 21 characters in body May25 asked An alternate proof of Egorov's Theorem