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You do not need $\sigma$-finiteness of the measure in Fubini theorem, although it is an hypothesis that can be assumed with no loss of generality, in that the support of an integrable function is, of …

In fact the total variation of a Banach valued vector measure is always a measure, as you are saying. What can happen is that $\|\mu\|$ has "heavy atoms", that is, measurable, non null sets $A\in\mat …

In fact, if $f$ is any measurable function on a measure space $(X,\mathcal{S},\mu)$, and wlog $f(x)>0$ on $X$, if for real numbers $a<b$ one has $\int_X f^a d\mu<+\infty$ and $\int_X f^b d\mu<+\inft …

If $(X,\Sigma, \mu)$ is semi-finite (it has no infinite atoms) and non-$\sigma$-finite, by transfinite induction there is a family $\{E_\alpha\}_{\alpha \in \omega_1}$ of disjoint measurable sets of …

We may rephrase the condition saying: $\mathcal{N}$ includes the class $\Sigma_0$ of all $E\in \Sigma$, such that $E$ is a null set for some non-atomic probability measure $\mu_E$ on $\Sigma$, or equ …

With the extra assumption it is true, and only continuity on $f:[0,\infty)\rightarrow [0,\infty)$ is needed. Of course, it is sufficient to show that some subsequence of $f(u_k)$ converges. So we can …

First point: indeed, without the extra assumption, $f(u_k)$ does not converge weakly to $f(u)$ in general, even if $u_k$ converges strongly. Take $u_k:=k \chi_ {\big[0,\frac{1}{f(k)}\big]}$. So $\|u_ …

Let e.g. $\nu$ be a finite Borel measure on $[a,b]$ and $m$ the Lebesgue measure. So the function $[a,b]\ni x\mapsto \nu\big([a,x)\big)$ is a BV function. It is therefore differentiable $m$-a.e., an …

No (unless X is a one-point space). The mean of two distinct shift-invariant product probability measures is a shift-invariant probability measure, though not a product.

Consider an uncountable discrete metric space $X $ (i.e., metrized by the Kronecker delta). Define a measure on $X$ putting for any $A\subset X,\\ $ $\mu(A)=1$ or $\mu(A)=0$ according whether $A$ belo …

If $(f_n)$ is a bounded sequence in $L^1(X,\mu)$, it is true that the set of $g\in L^\infty(X,\mu)$ such that $\int_X f_n g\, d\mu$ has a limit in $\mathbb{C}$ is a norm-closed linear subspace of $L^\ …

No. Let w.l.o.g $Q:= (0,1)$. There is a bounded linear functional $f$ on $E$ such that for any $x\in E$ one has: $\liminf _ {s\to 0} x(s)\le f(x) \le \limsup _ {s\to 0} x(s) $. This functional is posi …

I think you want the Pompeiu function (Feel free to add questions on the details of the construction).

A counterexample is a signed measure on the interval $I:=[-1,1]$ concentrated in the points $\{-1\}$, $\{0\}$, $\{1\}$ with weights respectively $1/2$, $-1$, $1/2$. (Thus $\int_If d\mu= f(1)/2+f(-1)/ …

For instance let $f:[0,1]\to[0,1]$ be the Cantor function and define $g(x):=x+f(x)$. Then $g:[0,1]\to[0,2]$ is a homeomorphism that maps the complement of the Cantor set $C$ onto a measure one open …