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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 ...
Bogdan's user avatar
  • 1,759
0 votes
0 answers
116 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(\...
Bogdan's user avatar
  • 1,759
5 votes
0 answers
140 views

Measure of the boundary of an BV-extension domain: do we have $|\nabla Eu|(\partial \Omega)=0?$

Let $\Omega\subset \Bbb R^d$ be open. The space $BV(\Omega)$ consists in functions $u\in L^1(\Omega)$ with bounded variation, i.e. $|u|_{BV(\Omega) }<\infty$ where \begin{align}\label{eq:bounded-...
Guy Fsone's user avatar
  • 1,101
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 ...
Xing Wang's user avatar