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Questions about abstract measure and Lebesgue integral theory. Also concerns such properties as measurability of maps and sets.
6
votes
1
answer
1k
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Tensor product of measure spaces
For a compact topological space $X$, denote by $\mathcal{M}(X)$ the Banach space of finite signed Borel (Radon) measures on $X$ with the total variation norm. This is canonically isometric to the dual …
42
votes
2
answers
3k
views
Measurability and Axiom of choice
In some situations, you need to show Lebesgue-measurability of some function on $\mathbb{R}^n$ and the verification is kind of lengthy and annoying, and even more so because measurability is "obvious" …
2
votes
2
answers
861
views
Uniformly bounded operator family and pointwise convergence
Let $1 \leq p < \infty$ be fixed and let $\Omega \subseteq \mathbb{R}^n$ be open. Let $(Q_n)_{n \in \mathbb{N}}$ be a uniformly bounded family of operators on $L^p(\Omega)$, i.e. there exists $C>0$ su …
1
vote
1
answer
271
views
Laplacian on space of measures
Let $X$ be a compact Riemannian manifold and let $\mathcal{M}(X)$ be the space of regular finite Borel measures with the total variation as norm.
The Laplace-Belrami-Operator $\Delta$ on $X$ with do …
15
votes
3
answers
6k
views
Dirac measures dense in space of measures? [closed]
Let $I$ be a compact interval and $\mathcal{M}(I)$ the space of (signed) Borel measures. We equip it with the weak topology, i.e. a sequence $\mu_n$ converges to zero if and only if
$$ \left|\int_I f( …
3
votes
1
answer
307
views
Pullback of $L^p$ functions via exponential map
Let $M$ be a complete Riemannian manifold, endowed with its exponential map $\exp: TM \longrightarrow M$. For any $C^k$- function $u$, we get the Pullback
$$ \exp^* u = u \circ \exp$$
which is in $C^k …
3
votes
1
answer
855
views
Measurability of subspace of set of all functions
Set $X=\mathbb{R}^n$ and let $X^{I}$, the space of maps from the (bounded or unbounded) interval $I$ to $X$, be endowed with the locally convex topology of pointwise convergence.
Is it true that the …
4
votes
3
answers
504
views
Bounded operators and axiom of choice
In the article below, it is shown that the proposition "Every linear operator defined on a whole Hilbert space is bounded" is consistent with the axioms of ZF + a weakened version of the axiom of choi …
2
votes
Difference between spaces of integrable functions w.r.t Lebesgue measure and Borel measure
I don't exactly know what the Lebesgue sigma-algebra is, but I presume you mean the extension of - for example - the Borel algebra that gives a complete measure. I know this as Baire algebra, and it h …