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1 answer
62 views

Integrability in the product space can follow from a property of the Nemytskii operator?

Let's say that $f:\Omega\times\mathbb{R}\to\mathbb{R}$ is a Caratheodory function (i.e. $f(x,\cdot)$ is continuous for a.a. $x\in\Omega$ and $f(\cdot,t)$ is measurable for all $t\in\mathbb{R}$), where ...
0 votes
0 answers
115 views

Integral of a measurable function with parameter is measurable?

Say that $f:\Omega\times\mathbb{R}\to\mathbb{R}$, where $\Omega\subset\mathbb{R}^N$ is an open set, is a function such that: $f(x,\cdot)\in L^1_{\text{loc}}(\mathbb{R})$ for a.a. $x\in\Omega$ $f(\...
0 votes
0 answers
66 views

Reference request: Integrability condition on measures

Let $(\mathcal{C}, \|\cdot\|)$ be a (non-locally compact) Banach space with Borel $\sigma$-algebra $\mathcal{B}$. Given a probability measure $\mu : \mathcal{B}\rightarrow[0,1]$, I'm interested in ...
2 votes
0 answers
162 views

$\int_{\mathbb{R}^{N}\setminus\Omega}\vert x-z\vert^{-N-\alpha} dz = c \ \forall x\in\partial U$ implies $dist(x,\partial\Omega)=c, x \in \partial U$?

Let $\alpha \in \mathbb R_+$, $\Omega \subset \mathbb R^N$ and $U \subset \Omega$. Is it true that if $$\int_{\mathbb R^N \setminus \Omega} |x - z|^{-N-\alpha} dz = \text{constant} \quad \text{for all ...
2 votes
0 answers
115 views

Showing that for measurable $\Omega \subseteq \mathbb{R}^n$, $L^1(\Omega; C_0(\mathbb{R}^n))$ is separable

Here we're integrating "Banach-valued" functions $u: \Omega \rightarrow C_0(\mathbb{R}^n))$ , and by $u \in L^1(\Omega; C_0(\mathbb{R}^n))$ I mean that $$\int_{x \in \Omega} \| u(x) \|_{\...
0 votes
1 answer
139 views

Change of variables for double integral [closed]

Thank you for your time. My basic question is whether the following change of variables allowed $$\int_0^a \int_0^b f(a-b)g(b-c)h(c)\,dc\,db = \int_0^a \int_0^b f(c)g(b-c)h(a-b)\,dc\,db$$ I fail to ...