Reputation
2,165
Next privilege 2,500 Rep.
Create tag synonyms
Badges
1 7 29
Newest
 Fanatic
Impact
~50k people reached

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.