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 May 30 comment Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$ @Shuhao Cao en.wikipedia.org/wiki/Absolute_continuity#Generalizations_2 also in Royden's Real Analysis May 30 revised Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$ added 192 characters in body May 30 revised Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$ added 4 characters in body May 30 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. May 30 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. May 30 revised Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$ deleted 9 characters in body May 30 asked Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$ May 29 accepted Rigorous proof that $\int_{\Omega}X\;dP=\int_{-\infty}^{\infty}xf(x)\;dx$ May 29 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. May 29 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 May 29 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. May 28 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 May 26 comment Rigorous proof that $\int_{\Omega}X\;dP=\int_{-\infty}^{\infty}xf(x)\;dx$ oh ok I see, thanks May 26 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 $X$ May 26 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? May 26 revised Rigorous proof that $\int_{\Omega}X\;dP=\int_{-\infty}^{\infty}xf(x)\;dx$ added 14 characters in body May 26 asked Rigorous proof that $\int_{\Omega}X\;dP=\int_{-\infty}^{\infty}xf(x)\;dx$ May 25 accepted An alternate proof of Egorov's Theorem May 25 comment An alternate proof of Egorov's Theorem you're right Norbert it isn't. May 25 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.