# Sargera

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 Sep6 comment Solution to Singular Integral Are you familiar with contour integration? Sep6 revised prove that $(\mathbb{R}^n,\|\cdot\|_\infty)$ is complete. added 157 characters in body Sep6 answered prove that $(\mathbb{R}^n,\|\cdot\|_\infty)$ is complete. Sep6 revised Averages of Indicator Functions of Compact Sets in $\mathbb{R}^{2}$ Over Circles added 660 characters in body Sep6 revised Averages of Indicator Functions of Compact Sets in $\mathbb{R}^{2}$ Over Circles added 316 characters in body Sep6 asked Averages of Indicator Functions of Compact Sets in $\mathbb{R}^{2}$ Over Circles Sep6 accepted Counter-Example (or Proof) to $\int_{0}^{1}f_{n}\;dx\to0$ Implies $f_{n}\to0$ a.e. $x$ Whenever $f_{n}\geq0$. Sep6 revised Limit of Integral of Difference Quotients of Measurable/Bounded $f$ Being $0$ Implies $f$ is Constant added 97 characters in body Sep6 comment Counter-Example (or Proof) to $\int_{0}^{1}f_{n}\;dx\to0$ Implies $f_{n}\to0$ a.e. $x$ Whenever $f_{n}\geq0$. You're right. I think I need to take a second and rethink, since actually in my comment I really want $\lim nf_{n}$ to converge to $0$ a.e. $x$, not just $f_{n}$. Sep6 comment Counter-Example (or Proof) to $\int_{0}^{1}f_{n}\;dx\to0$ Implies $f_{n}\to0$ a.e. $x$ Whenever $f_{n}\geq0$. Ah. Typewriter sequence. So what if we add the auxiliary condition that $\int_{0}^{1}f_{n}\;dx\to0$ "rapidly" in the sense that $n\int_{0}^{1}f_{n}\;dx\to0$ as $n\to\infty$? Not that the counter-example exploited possible unboundedness of $f_{n}$, but also assume $0\leq f_{n}\leq M$ along with the rapid convergence condition. This is the condition of the convergence in the question I linked in the original post. Sep6 revised Counter-Example (or Proof) to $\int_{0}^{1}f_{n}\;dx\to0$ Implies $f_{n}\to0$ a.e. $x$ Whenever $f_{n}\geq0$. edited title Sep6 revised Limit of Integral of Difference Quotients of Measurable/Bounded $f$ Being $0$ Implies $f$ is Constant added 220 characters in body Sep6 asked Counter-Example (or Proof) to $\int_{0}^{1}f_{n}\;dx\to0$ Implies $f_{n}\to0$ a.e. $x$ Whenever $f_{n}\geq0$. Sep6 comment Limit of Integral of Difference Quotients of Measurable/Bounded $f$ Being $0$ Implies $f$ is Constant @Jonathan I incorporated your comments in the most recent edit. Sep6 revised Limit of Integral of Difference Quotients of Measurable/Bounded $f$ Being $0$ Implies $f$ is Constant added 1133 characters in body Sep5 comment Limit of Integral of Difference Quotients of Measurable/Bounded $f$ Being $0$ Implies $f$ is Constant Thank you for making me realize my naivety. Approximations to the identity much... (: Sep5 comment Limit of Integral of Difference Quotients of Measurable/Bounded $f$ Being $0$ Implies $f$ is Constant Isn't this true though when $g\geq0$? How can you have $\int_{0}^{1}g(x)\;dxM$ on a set of positive measure? Sep5 comment Limit of Integral of Difference Quotients of Measurable/Bounded $f$ Being $0$ Implies $f$ is Constant math.ucla.edu/grad/handbook/hbquals.shtml -- See problem #1 from Analysis Spring 2013. I don't understand your point though; there are no singularities or otherwise obstructions to speak of that would prevent us from integrating along $[0,1]$... Sep5 asked Limit of Integral of Difference Quotients of Measurable/Bounded $f$ Being $0$ Implies $f$ is Constant Jul19 answered $f^{-1}$ is continuously differentiable.