functional analysis – Can we use the duality notation such that the second variable is an element of the measure space?

I read article New Sequential Compactness Results for Spaces of Scalarly Integrable Functions by Erik J. Balder, on page 8 : Author defined the function $a: Ttimes Eto mathbb{R}$ by the usual duality between $E$ and $E^*=F$ $( ⟨.,.⟩)$ such that:
$$
a(t,x)=langle x, trangle
$$

But normally the second variable $(tin T)$ must be an element of dual of $E£, right? Why does the author make this notation?

An idea please
enter image description here

measure theory – Null preserving transformation

Suppose that $(Omega,mu)$ is a measure space. Let $tau:OmegatoOmega$ is a measurable map such that $mucirctau^{-1}<<mu$. Then $tau$ s said to be null preserving. I want to prove the following. If $f:Omegatomathbb R$ is measurable and $mu({fneq fcirc f})=0,$ then there exists a measurable function $f^prime$ such that $f^prime=f^prime circ tau$ and $mu(fneq f^prime)=0.$ If we define $A:={xinOmega:f(x)=f(tau(x))}.$ I can prove that $A$ is $tau$-invariant mod $mu.$ A natural way to define $f^prime$ would be $f^prime=f1_{B}$ where $B$ is $tau$-invariant and $mu(ADelta B)=0.$ But I can not really see if it works. It will work definitely if we have $Bsubseteq A.$ We can have $B$ to be the set $cup_{k=0}^infty(Asetminus cup_{k=0}^inftytau^{-k}(Asetminus tau^{-1}A)).$ Can we have that $Bsubseteq A$? Also. I want to find some intuitive idea how the construction should be.

Will it be the most humiliating day of every libs life if studies prove that hydroxicholoroquine works as a preventative measure?

That is not the way that sensible people handle things. 

. The question is not whether it works as a preventative measure, as much as whether it is a relatively safe chemical/drug to take..If it was suggested that swallowing battery acid might work, it is a bad idea even if it did kill the virus or the common cold or anything..I think you are just hoping for a little more drama  

fa.functional analysis – $ frac{1}{k} sum_{i = 1}^{k} {g_{i}} to g_{infty}$ weakly a.e, imply that $frac{1}{k} sum_{i = 1}^{k} {g_{i}} to g_{infty} $ in measure?

Let $(E,mathcal{A},mu)$ be a finite measure space and $(X,|.|)$ be a reflexive Banach space. The en set of all Bochner-integrable function from $E$ to $X$ is denoted by $mathcal{L}_{X}^{1}$.

$(i)$ $F subset mathcal {L}^0_X $ is a closed in measure if $ F $ is empty or any sequence $ {f_n } $ of elements of $ F $ which converges in measure to an element $ f in mathcal{L}^0_X $ has its limit $ f $ in $ F $.

$(ii)$ $ phi: mathcal{L}^0_X to overline {mathbb{R}} $ is said to be lower semi-continuous in measure if:
$$
forall a in mathbb {R} ~: ~ {f in mathcal {L}^0_X ~: ~ phi (f) leq a } text {is a closed in measure}.
$$

Theorem:

Suppose that $I:mathcal{L}_{X}^{1}to (-infty,+infty)$ is a convex functional which is lower semi-continuous in measure and that $Bin mathcal{L}_{X}^{1}$ is convex, closed in measure and uniformly bounded in $mathcal{L}^{1}$-norm.

Show that $I$ attains its minimum on $B$.

Hint. Apply the following theorem :

Theorem:

Suppose that $ (f_n)_{ngeq 1} subset mathcal {L}_{X}^1$ is a seqsequence with : $$sup_n int_{E}{|f_n| dmu} < infty .$$Then there exist $ h _{infty} in mathcal {L}_{X}^1 $ and a subsequence $ (g_k)_k $ of $(f_n)_n $ such that for every subsequence $ (h_m)_m $ of $(g_k)_k$ : $$ frac{1}{n}sum_{j=1}^{n}{h_j}underset{n}{to} h _{infty} ~~~~~~text{weakly in}~X~~text{ a.e.}$$

My effort:

We set $ {displaystyle m: = inf_ {f in B} I (f)} $. Suppose that $ m <+ infty $, according to the definition of the lower bound, we have:
$$
forall n in mathbb{N} ~,~ exists f_n in B text{ such as: } I (f_n) <m + frac{1}{n}
$$

as $ B $ is uniformly bounded in $ mathcal{L}_{X}^1$, in particular, we have:
$$
sup_n int_{E} {| f_n | d mu} <infty.
$$

According to the previous theorem, there exists $ g_infty in mathcal{L}_{X}^1 $ and a subsequence $ {g_n} $ of $ {f_n } $ such that:
$$
frac{1}{k} sum_{i = 1}^{k} {g_{i}} overset {sigma (X, X^{*})} {longrightarrow} g_{infty} qquad a.e
$$

therefore $ {displaystyle frac{1}{k} sum_{i = 1}^{k} {g_ {i}}} $ converges in measure to $ g_{infty} $.
Since $ B $ is convex therefore:
$$
frac{1}{k} sum_{i = 1}^{k} {g_ {i}} in B
$$

$ B $ is closed in measure therefore:
$$
g_{infty} in B
$$

as $ I $ is semi-continuous inferior in measure and convex, therefore:
$$
I (g_{infty}) leq liminf_{n} I (frac{1}{k} sum_{i=1}^{k}{g_{i}}) leq liminf_{n} frac{1}{k} sum_{i=1}^{k}{I (g_{i})} qquad text{ en mesure }
$$

so :
$$
I (g_{infty}) leq liminf_{k} frac{1}{k} sum_ {i = 1}^{k} {big (m + frac{1}{i} big) } = m + liminf_{k} frac {sum_{i = 1}^{k}{frac{1}{i}}}{k} leq m + liminf_{k} frac {1+ log(k)}{k}
$$

like $ {displaystyle liminf_{k} frac {1+ log (k)}{k} = 0} $, so $ I (g _ {infty}) leq m $. from where :
$$
I (g_{infty}) = inf_{f in B} I (f).
$$

My problem: Do I have the right to say that:$ frac{1}{k} sum_{i = 1}^{k} {g_{i}} overset {sigma (X, X^{*})} {longrightarrow} g_{infty}$ a.e, imply that $frac{1}{k} sum_{i = 1}^{k} {g_{i}} $ converges in measure to $ g_{infty} $? If the answer is no, please have an idea.

functional analysis – Measure theory: motivation behind monotone convergence theorem

I am watching a very nice set of videos on measure theory, which are great. But I am not clear on what the motivation is behind the monotone convergence theorem–meaning why we need it?

The statement of the theorem is that given a set of functions $f_n$ such that $f_1 leq f_2 leq f_3 leq … f_n leq f$

$$
lim_{n rightarrow infty} int_X f_n dmu = int_X lim_{n rightarrow infty } f_n dmu = int_X f dmu
$$

So the theorem suggests the interchange of the limit and the integral sign. But I am not sure what the implications of this interchange is and under what conditions this interchange is not possible (for the Lebesgue integral)? Meaning, that this particular theorem of monotone convergence presupposes the Lebesgue integral as opposed to the Riemann integral. So, is monotone convergence not guaranteed for the Riemann integral–is that the key distinction? And second, are the cases where monotone convergence fails for the Riemann integral due to the fact that the Riemann integral gives mass to sets in $X$ of measure zero, while the Lebesgue integral does not have this problem?

probability – Calculating the quotient $ frac { mu (E)} { nu (E)} $ of two measures with density w.r.t. the Lebesgue measure

Suppose that $ mu $ and $ nu $ are probability measures on $ mathbb {R} $ with density $ f $ and $ g $ w.r.t. the Lebesgue measure, i.e. $ mu = fdx $ and $ nu = g dx $. Is there an easy way to calculate the qutotient
$$
frac { mu (E)} { nu (E)}
$$

of a measurable set $ E $? This is a quotient of two integrals of $ f $ and $ g $. Can we calculate one single integral over the quotient of $ f $ and $ g $ instead?

Does the below-quoted fact follow from the countable subadditivity property of probability measure $mathbb{P}$?

Let $(Omega,mathcal{F},mathbb{P})$ be a probability space, $a$ and $b$ be two rationals (i.e. $a$,$b$ $in mathbb{Q}$) such that $a<b$ and $(X_n)$ be a sequence of random variables defined on the above-defined measurable space. Set:
begin{equation}
Lambda_{a,b}={limsuplimits_{nrightarrowinfty}X_ngeq b; liminflimits_{nrightarrowinfty}X_n leq a}
end{equation}

begin{equation}
Lambda=bigcuplimits_{a<b} Lambda_{a,b}
end{equation}

Therefore, I “read” $Lambda$ as follows:

there exists at least a pair of rationals $a$, $b$ ($a<b$) such that $limsuplimits_{nrightarrowinfty}X_ngeq b; liminflimits_{nrightarrowinfty}X_n leq a$, that is

begin{equation}
Lambda={limsuplimits_{nrightarrowinfty}X_n> liminflimits_{nrightarrowinfty}X_n}
end{equation}

I am given that $mathbb{P}(Lambda_{a,b})=0$ $forall a, b in mathbb{Q}$ such that $a<b$. And, at this point, I know that one can state that

$mathbb{P}(Lambda)=0$, since all rational pairs are countable”

I interpret the above statement in the following way, but I am not sure whether it is correct or not.


Since:

  • $mathbb{P}(Lambda_{a,b})=0$ $forall a,b in mathbb{Q}$ such that $a<b$;
  • $bigcuplimits_{a<b}Lambda_{a,b}$ is a countable union, since rationals are
    countable by definition;

by countable subadditivity property of probability measure $mathbb{P}$, it follows that:
begin{equation}
mathbb{P}(Lambda)=mathbb{P}big(bigcuplimits_{a<b} Lambda_{a,b}big)leq sumlimits_{a<b}mathbb{P}big(Lambda_{a,b}big)=0
end{equation}

where $sumlimits_{a<b}mathbb{P}big(Lambda_{a,b}big)=0$ follows from the fact that I am given that $mathbb{P}(Lambda_{a,b})=0$ $forall a, b in mathbb{Q}$ such that $a<b$.

Hence,since probability lies by definition between $0$ and $1$, $mathbb{P}(Lambda)leq 0$ “means” that:
begin{equation}
mathbb{P}(Lambda)=0
end{equation}


Is my reasoning correct?

measure theory – Does $limint f_n=int f$ imply $limint|f_n|=int|f|$?

Suppose that ${f_n}$ is a sequence complex measurable functions on a measurable space $(X,Omega,mu)$.
Let $f$ be the pointwise limit of $f_n$.
Does (1) implies (2)? where
$$lim_{ntoinfty}int_Xf_n,dmu=int_Xf,dmu.tag{1}$$
$$lim_{ntoinfty}int_X|f_n|,dmu=int_X|f|,dmu.tag{2}$$

(EDIT) I don’t know if the above statement is valid or not.
So please give me a counterexample if it is false.

My trial:
(1) implies
$$lim_{ntoinfty}int_X(f_n-f),dmu=0.tag{3}$$
$$lim_{ntoinfty}int_X|f_n-f|,dmu=0.tag{4}$$
$$lim_{ntoinfty}int_XBig||f_n|-|f|Big|,dmu=0.tag{5}$$
$$lim_{ntoinfty}int_X|f_n|-|f|,dmu=0.tag{6}$$
So (2) follows.
I’m not certain whether (3) implies (4).
That is to say that (7) implies (8) where $g_nto0$ and
$$lim_{ntoinfty}int_Xg_n,dmu=0.tag{7}$$
$$lim_{ntoinfty}int_X|g_n|,dmu=0.tag{8}$$
Let $A_n={xin X:|u_n(x)|ge |v_n(x)|}$ and $B_n=Xsetminus A_n$.
Then
$$int_X|g_n|,dmu
=int_Xsqrt{(u_n(x))^2+(v_n(x))^2},dmu
leint_{A_n}sqrt2|u_n|,dmu+int_{B_n}sqrt2|v_n|,dmu
lesqrt2int_X|u_n|,dmu+sqrt2int_X|v_n|,dmu$$

So, it is enough to consider “(7) implies (8)” for real $g_n$.

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