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6h
comment Why is this relation transitive?
Using the = sign when you really mean "implies" is confusing.
6h
revised Opinions on Lax's “Calculus with Applications, 2e”
mechanics
7h
comment Convergence in measure theory
Are you working in a finite measure space?
1d
comment measurability of supremum of a class of functions
By the way, if $y \in N$ was a typo and you meant $(x,y) \in N$, then the function f you get is Borel, but $f \circ h \ne I(M)$.
2d
comment measurability of supremum of a class of functions
@danielson: Indeed, I think this answer is answering a different question than the one that was asked.
2d
comment measurability of supremum of a class of functions
Indeed, a composition of Lebesgue measurable functions is not in general Lebesgue measurable (so my argument doesn't contradict your example). (See this answer of mine on MO for further details.) But a composition of Borel functions is definitely Borel; this is immediate from the definition. I don't understand the example in your comment: $N$ is a subset of $\mathbb{R}^2$ but $y \in \mathbb{R}$ so writing $I(y \in N)$ doesn't make sense.
2d
comment Is a linear operator on $\ell^2$ defined by the inner product necessarily bounded?
possible duplicate of If $\sum a_n b_n <\infty$ for all $(b_n)\in \ell^2$ then $(a_n) \in \ell^2$. See also math.stackexchange.com/questions/58565/…
Jul
27
comment Strategy verifying Riemann Hypothesis?
What is your question?
Jul
26
comment Expectation of the absolut value of the determinant of a random matrix
@Count: That's a long document. Can you point us to the relevant part, and maybe briefly summarize what's to be found there?
Jul
26
comment Finite additivity in outer measure
@user160307: The union of the translates is still a subset of [0,1]. We don't need to compute its outer measure exactly; by monotonicity it is certainly at most 1.
Jul
26
answered Finite additivity in outer measure
Jul
26
comment Show that different eigenfunctions of integral kernel are orthogonal
You will first need to verify that $K$ is self-adjoint, i.e. $(Ku,v)=(u,Kv)$.
Jul
26
comment measurability of supremum of a class of functions
Define $h : [0,1] \to [0,1]^2$ by $h(x)=(x,x)$, which is continuous, hence Borel. Then $f \circ h= I_M$ which is not Borel.
Jul
26
comment Integration respect the Lévy measure
Those are two different notations for the same thing.
Jul
25
comment Trace theorem for Sobolev functions: what is the significance of continuous extension to the boundary?
For example, if you drop condition 1, there is no guarantee that T is at all a reasonable notion of "take the boundary value". Taking T=0 would satisfy condition 2.
Jul
25
answered Alternative proof for the fact that a continuous function on a closed interval attains its boundaries.
Jul
25
comment measurability of supremum of a class of functions
I guess this depends on what is meant by "measurable". Your function $f$ is Lebesgue measurable but not Borel.
Jul
25
comment measurability of supremum of a class of functions
@user79963: What measurable set are you talking about? Are you wondering why $f$ is measurable? Note that $f = 1_A$ where $A = \{(y,y) : x \in Y\}$. You can easily check that $A$ is actually closed in $X \times Y$, and in particular measurable.
Jul
25
comment Can we expect to choose $ f\in L^{1}(\mathbb R)$ such that $\|f\|_{L^{1}(\mathbb R)}\leq 1,$ and $\hat{f}(x_{0})=0$?
You can even get $\|f\|_{L^1} = 1$. Let $g$ be some very nice nonzero function (say $C^\infty_c$) with $g(x_0) = 0$ and let $f$ be its inverse Fourier transform. Since $g$ is very smooth we have $f \in L^1$ and clearly $\|f\|_{L^1} \ne 0$. Now rescale.
Jul
24
comment Does differentiability imply absolute continuity?
For future reference, when you find a question that has already been answered, rather than posting an answer with a link, it is better to use the "flag" button to indicate that the question is a duplicate.