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Mar
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awarded  Enlightened
Mar
4
awarded  Nice Answer
Feb
20
accepted If $f$ is bounded non-negative $L^1$, is $f\leq g$ a.e. for some continuous integrable $g$?
Feb
18
comment If $f$ is bounded non-negative $L^1$, is $f\leq g$ a.e. for some continuous integrable $g$?
@A.S. could you add some details on how to construct the transitions while ensuring $g\geq f$?
Feb
18
comment If $f$ is bounded non-negative $L^1$, is $f\leq g$ a.e. for some continuous integrable $g$?
Wlog $0< f <1$, in which case $E_1=\mathbb{R}$ has infinite measure. I think indeed something like Lusin's theorem may be needed, but I like your idea of approximating by open sets.
Feb
18
asked If $f$ is bounded non-negative $L^1$, is $f\leq g$ a.e. for some continuous integrable $g$?
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7
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Jan
25
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25
awarded  Notable Question
Jan
24
awarded  Nice Answer
Jan
18
answered Does free bimodule exist?
Dec
16
awarded  Popular Question
Dec
8
awarded  differential-geometry
Dec
6
awarded  Enlightened
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awarded  Nice Answer
Nov
16
comment Prove or Disprove: if $\lim\limits_{n \to \infty} (a_{2n}-a_n)=0,$ then $\lim\limits_{n \to \infty} a_n=0.$
What about the constant sequence $a_n=1$?
Nov
8
comment First-order definition of “$f$ is continuous at $x$” using just $<$
@avid19 I see, is that the problem, that there are no such symbols in the language? But how is $\Bbb R$ to be understood in this context?
Nov
8
comment First-order definition of “$f$ is continuous at $x$” using just $<$
Wouldn't something like $$\varphi(x)=\forall\epsilon >0~\exists\delta>0~\forall y~ (x-\delta<y<x+\delta\rightarrow f(x)-\epsilon<f(y)<f(x)+\epsilon)$$ do the trick?
Sep
27
comment Is the homotopy category cartesian closed?
If you consider unbased paths, then there is a homotopy equivalence $X\simeq X^I$.
Sep
27
comment Is the homotopy category cartesian closed?
@MikeMiller ok, but why then does the OP say that the path space is isomorphic to the discrete set of path components? This doesn't seem to gel with a basepoint approach.