Explore Measurement Concepts with Our Examples and Questions

Why does $X\in <!--\; \in \; -->{\mathcal{L}}^p(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{F},\mathbb{P}),p\in <!--\; \in \; -->[1,\mathrm{\infty <!--\; \infty \; -->})$, implies that $\mathbb{E}[X|\mathcal{A}]\in <!--\; \in \; -->{\mathcal{L}}^p(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{A},\mathbb{P})$, where $\mathcal{A}\subset <!--\; \subset \; -->\mathcal{F}$ is a sub-$\mathrm{\sigma <!--\; \sigma \; -->}$-algebra?

The defintion of the conditional expectation says that $\mathbb{E}[X{\mathbb{1}}_A]=\mathbb{E}[\mathbb{E}[X|\mathcal{A}]{\mathbb{1}}_A]$ for all $A\in <!--\; \in \; -->\mathcal{A}$ but why can we conclude from this that $\mathbb{E}[|X{|}^p{\mathbb{1}}_A]=\mathbb{E}[\mathbb{E}[|X{|}^p|\mathcal{A}]{\mathbb{1}}_A]$ for all $A\in <!--\; \in \; -->\mathcal{A}$?

What is the best way to convert variance in one measurement unit to variance in another measurement unit? For example:
The variance of the heights of the mango trees in your garden is 13 square feet. What will be the variance in yards?

Suppose $E\subset <!--\; \subset \; -->{\mathbb{R}}^d$ and ${\mathcal{O}}_n$ is the open set
$\begin{array}{r}{\mathcal{O}}_n=\{x\in <!--\; \in \; -->{\mathbb{R}}^d:d(x,E)<\frac{1}{n}\}\end{array}$
If $E$ is compact, then $m(E)=\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}m\left({\mathcal{O}}_n\right)$. I wonder what are some examples that this equality does not hold when $E$ is closed and unbounded, or $E$ is open and bounded.

Given $\mathrm{\Omega <!--\; \Omega \; -->}=(0,1)$, $\mathcal{A}$ is the Borel Sigma-Algebra restricted to $\mathrm{\Omega <!--\; \Omega \; -->}$, and $\mathrm{\mu <!--\; \mu \; -->}$ is the Lebesgue measure, let $f(\mathrm{\omega <!--\; \omega \; -->})=\frac{1}{\mathrm{\omega <!--\; \omega \; -->}}$ and
$0<\mathrm{\omega <!--\; \omega \; -->}<\frac{1}{n}$
$\frac{1}{n}\le <!--\; \le \; -->\mathrm{\omega <!--\; \omega \; -->}<1$
I have to prove that $f_n$ converges $\mathrm{\mu <!--\; \mu \; -->}$-almost uniformly to $f$.
From what I learned in class, I have to show:
$\mathrm{\forall <!--\; \forall \; -->}\mathrm{ϵ<!--\; ϵ\; -->}>0,\mathrm{\exists <!--\; \exists \; -->}{E}_{\mathrm{ϵ<!--\; ϵ\; -->}}\in <!--\; \in \; -->\mathcal{A},\mathrm{\mu <!--\; \mu \; -->}(E_{\mathrm{ϵ<!--\; ϵ\; -->}}^c)<\mathrm{ϵ<!--\; ϵ\; -->}⟹<!--\; ⟹\; -->f_n\text{ converges uniformly to }f\text{ on }{E}_{\mathrm{ϵ<!--\; ϵ\; -->}}.$
My idea was to let $E_{\mathrm{ϵ<!--\; ϵ\; -->}}=[\mathrm{ϵ<!--\; ϵ\; -->}-<!--\; -\; -->0.000000001,1)$. Then $E_{\mathrm{ϵ<!--\; ϵ\; -->}}^c=(0,\mathrm{ϵ<!--\; ϵ\; -->}-<!--\; -\; -->0.000000001)⟹<!--\; ⟹\; -->\mathrm{\mu <!--\; \mu \; -->}(E_{\mathrm{ϵ<!--\; ϵ\; -->}}^c)<\mathrm{ϵ<!--\; ϵ\; -->}.$
To prove uniform convergence, I need:
$\mathrm{\forall <!--\; \forall \; -->}\mathrm{\delta <!--\; \delta \; -->}>0,\mathrm{\exists <!--\; \exists \; -->}{N}_{\mathrm{\delta <!--\; \delta \; -->}}⟹<!--\; ⟹\; -->\mathrm{\forall <!--\; \forall \; -->}n>N_{\mathrm{\delta <!--\; \delta \; -->}},\mathrm{\forall <!--\; \forall \; -->}\mathrm{\omega <!--\; \omega \; -->}\in <!--\; \in \; -->E_{\mathrm{ϵ<!--\; ϵ\; -->}},|f_n(\mathrm{\omega <!--\; \omega \; -->})-<!--\; -\; -->f(\mathrm{\omega <!--\; \omega \; -->})|<\mathrm{\delta <!--\; \delta \; -->}.$
Since $E_{\mathrm{ϵ<!--\; ϵ\; -->}}=[\mathrm{ϵ<!--\; ϵ\; -->}-<!--\; -\; -->0.000000001,1),$ $\mathrm{\omega <!--\; \omega \; -->}\ge <!--\; \ge \; -->\mathrm{ϵ<!--\; ϵ\; -->}-<!--\; -\; -->0.000000001$, so if $\mathrm{ϵ<!--\; ϵ\; -->}-<!--\; -\; -->0.000000001\ge <!--\; \ge \; -->\frac{1}{n}$, we get $|f_n(\mathrm{\omega <!--\; \omega \; -->})-<!--\; -\; -->f(\mathrm{\omega <!--\; \omega \; -->})|=0<\mathrm{\delta <!--\; \delta \; -->}.$
This holds for all $n\ge <!--\; \ge \; -->\frac{1}{\mathrm{ϵ<!--\; ϵ\; -->}-<!--\; -\; -->0.000000001}$, so we let $N_{\mathrm{\delta <!--\; \delta \; -->}}=\frac{1}{\mathrm{ϵ<!--\; ϵ\; -->}-<!--\; -\; -->0.000000001}$, proving the almost uniform convergence.
Is this argument correct?

Let $M:=(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{F})$ be a measurable space and let $\mathcal{M}$ be the vector space of all $(\mathbb{R},\mathcal{B}(\mathbb{R}))$-valued measurable functions defined on $M$. If $G$ is a sub-$\mathrm{\sigma <!--\; \sigma \; -->}$-field of $\mathcal{F}$, let ${\mathcal{M}}_G$ be the subspace of $\mathcal{M}$ whose elements are the $G$-measurable functions.
Suppose that ${\mathcal{F}}_1,\dots <!--\; \dots \; -->,{\mathcal{F}}_k$ are sub-$\mathrm{\sigma <!--\; \sigma \; -->}$-fields of $\mathcal{F}$ and let $\mathfrak{F}:={\vee <!--\; \vee \; -->}_{i=1}^k{\mathcal{F}}_i$ denote their join.
Is there any relationship between ${\mathcal{M}}_{\mathfrak{F}}$ and ${\mathcal{M}}_{{\mathcal{F}}_i}$ besides that the latter is a subset of the former? Something like the former being the sum of the latter?
$\underset{i=1}{\overset{k}{⨁<!--\; ⨁\; -->}}{\mathcal{M}}_{{\mathcal{F}}_{\mathcal{i}}}\subseteq <!--\; \subseteq \; -->{\mathcal{M}}_{\mathfrak{F}}$ holds (where $\underset{i=1}{\overset{k}{⨁<!--\; ⨁\; -->}}{\mathcal{M}}_{{\mathcal{F}}_{\mathcal{i}}}:=\{f_1+\dots <!--\; \dots \; -->+f_k:f_i\in <!--\; \in \; -->{\mathcal{M}}_{{\mathcal{F}}_i},i\in <!--\; \in \; -->\{1,\dots <!--\; \dots \; -->,k\}\}$, but I can't see how to prove the reverse inclusion if it is indeed true.

Let $(E_{\mathrm{\alpha <!--\; \alpha \; -->}}{)}_{\mathrm{\alpha <!--\; \alpha \; -->}\in <!--\; \in \; -->I}$, where $I$ is an index set (for example, [0,1]). Can we define the corresponding liminf, for example,
$\underset{\mathrm{\alpha <!--\; \alpha \; -->}\searrow <!--\; \searrow \; -->0}{lim inf}{E}_{\mathrm{\alpha <!--\; \alpha \; -->}}=\underset{r>0}{\bigcup <!--\; \bigcup \; -->}\underset{0<\mathrm{\alpha <!--\; \alpha \; -->}<r}{\bigcap <!--\; \bigcap \; -->}{E}_{\mathrm{\alpha <!--\; \alpha \; -->}}.$
My desired result is the following continuous parameter type Fatou's lemma: Let $\mathrm{\mu <!--\; \mu \; -->}$ be a finite meaure and each $E_{\mathrm{\alpha <!--\; \alpha \; -->}}$ is measurable, then
$\mathrm{\mu <!--\; \mu \; -->}(\underset{\mathrm{\alpha <!--\; \alpha \; -->}\searrow <!--\; \searrow \; -->0}{lim inf}{E}_{\mathrm{\alpha <!--\; \alpha \; -->}})\le <!--\; \le \; -->\underset{\mathrm{\alpha <!--\; \alpha \; -->}\searrow <!--\; \searrow \; -->0}{lim inf}\mathrm{\mu <!--\; \mu \; -->}(E_{\mathrm{\alpha <!--\; \alpha \; -->}}).$
I am not sure whether these make sense. Any comments are welcome! Many thanks!

I'm working through some literature from Geometric Measure Theory for an assignment paper and have got stuck at a step in the proof of the Federer-Volpert theorem. The reasoning is as follows: according to the BV coarea formula, the level sets $\{u>t\}$ have finite perimeter for $\mathrm{\lambda <!--\; \lambda \; -->}$-a.e. $t\in <!--\; \in \; -->\mathbb{R}$. From this, the author concludes that we can find a dense subset $D$ of $\{t\in <!--\; \in \; -->\mathbb{R}|\{u>t\}\text{has finite perimeter}\}$. I'm guessing this must follow from some well-known fact in measure theory, but I can't put my finger on it. Any help would be greatly appreciated.

We can consider the sequence space ${\mathbb{R}}^{\mathbb{Z}}$, which becomes a Polish space (complete, separable metric space) when equipped with the product topology. In the theory of stochastic processes, one is frequently interested in probability measures on this space.
A classical result on measure theory in Polish spaces tells us that any probability measure $\mathrm{\mu <!--\; \mu \; -->}$ on a Polish space $X$ is tight: given $\mathrm{ϵ<!--\; ϵ\; -->}>0$, then there is a compact set $K_{\mathrm{ϵ<!--\; ϵ\; -->}}\subseteq <!--\; \subseteq \; -->X$ such that $\mathrm{\mu <!--\; \mu \; -->}(K_{\mathrm{ϵ<!--\; ϵ\; -->}})>1-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->}$.
Hence any product probability measure on ${\mathbb{R}}^{\mathbb{Z}}$ is tight.
Say we fix an i.i.d. probability measure $\mathrm{\mu <!--\; \mu \; -->}$ on ${\mathbb{R}}^{\mathbb{Z}}$. Precisely, $\mathrm{\mu <!--\; \mu \; -->}(\prod <!--\; \prod \; -->A_i)=\prod <!--\; \prod \; -->\mathrm{\nu <!--\; \nu \; -->}(A_i)$ for some probability measure $\mathrm{\nu <!--\; \nu \; -->}$ on $\mathbb{R}$.
Is there a nice, concrete proof that $\mathrm{\mu <!--\; \mu \; -->}$ is tight (independent of $\mathrm{\nu <!--\; \nu \; -->}$)?
The reason I ask is tightness is somewhat counter-intuitive here. For instance, if $\mathrm{\nu <!--\; \nu \; -->}$ has unbounded support, then, for each $M>0$, the set $\{k\in <!--\; \in \; -->\mathbb{Z}\mid <!--\; \mid \; -->|x_k|\ge <!--\; \ge \; -->M\}$ is infinite almost surely under $\mathrm{\mu <!--\; \mu \; -->}$ (by the 0-1 law). Thinking along those lines, it's not easy to see $K_{\mathrm{ϵ<!--\; ϵ\; -->}}$ intuitively.

I am trying to solve the following problem:
Let $(X,S,\mathrm{\mu <!--\; \mu \; -->})$ be a measure space. Let $1<p<\mathrm{\infty <!--\; \infty \; -->}$. Let $f:X\times <!--\; \times \; -->X\to <!--\; \to \; -->\mathbb{R}$ be such that, for every $y\in <!--\; \in \; -->X$, the section $f^y$ is $p$-integrable and that
${\displaystyle {\int <!--\; \int \; -->}_X\Vert <!--\; \Vert \; -->f^y{\Vert <!--\; \Vert \; -->}_{p}d\mathrm{\mu <!--\; \mu \; -->}(y)<+\mathrm{\infty <!--\; \infty \; -->}.}$
Define, for $x\in <!--\; \in \; -->X$,
$g(x)={\displaystyle {\int <!--\; \int \; -->}_{X}f(x,y)d\mathrm{\mu <!--\; \mu \; -->}(y).}$
Show that $g\in <!--\; \in \; -->L^p(\mathrm{\mu <!--\; \mu \; -->})$ and that
$\Vert <!--\; \Vert \; -->g{\Vert <!--\; \Vert \; -->}_p\le <!--\; \le \; -->{\displaystyle {\int <!--\; \int \; -->}_X\Vert <!--\; \Vert \; -->f^y{\Vert <!--\; \Vert \; -->}_{p}d\mathrm{\mu <!--\; \mu \; -->}(y).}$"

So above is the problem. We need to show (after simplifying the inequality) that
${\left({\displaystyle {\int <!--\; \int \; -->}_X{|{\int <!--\; \int \; -->}_{X}f(x,y)d\mathrm{\mu <!--\; \mu \; -->}(y)|}^{p}d\mathrm{\mu <!--\; \mu \; -->}(x)}\right)}^{\frac{1}{p}}\le <!--\; \le \; -->{\displaystyle {\int <!--\; \int \; -->}_X{({\int <!--\; \int \; -->}_X{|f(x,y)|}^{p}d\mathrm{\mu <!--\; \mu \; -->}(x))}^{\frac{1}{p}}d\mathrm{\mu <!--\; \mu \; -->}(y).}$
Can you please give me some hints on how to do it? Thanks.

The purpose of this problem is that I want to prove that for any $\mathrm{\lambda <!--\; \lambda \; -->}$ integrable function f on a bounded closed interval [a,b] holds
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\int <!--\; \int \; -->}_{[a,b]}f(x)\sin <!--\; \; -->(nx)d\mathrm{\lambda <!--\; \lambda \; -->}=0.$
I have submitted a proof below.

a saw some proofs of Lebesgue regularity and open set G (for E$\in <!--\; \in \; -->\mathcal{L}$) such that $\mathrm{\lambda <!--\; \lambda \; -->}(G\setminus <!--\; \setminus \; -->E)<\mathrm{\epsilon <!--\; \epsilon \; -->}$, but how to proof this one: Namely suppose $E\in <!--\; \in \; -->\mathcal{L}$ is such that for $\mathrm{\epsilon <!--\; \epsilon \; -->}>0$ there exist a closed set $F\subset <!--\; \subset \; -->E$ such that $\mathrm{\lambda <!--\; \lambda \; -->}(E\setminus <!--\; \setminus \; -->F)<\mathrm{\epsilon <!--\; \epsilon \; -->}$?

I'm wondering whether the following statement holds:
Let $f_n,f:<!--\; :\; -->\mathbb{R}\to <!--\; \to \; -->{\mathbb{R}}_0^{+}$ be functions with $\int <!--\; \int \; -->f_n(x)dx=\int <!--\; \int \; -->f(x)dx=1$ and for every bounded and countinuous function $g:<!--\; :\; -->\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ the following integral-convergence
$\int <!--\; \int \; -->g(x)\cdot <!--\; \cdot \; -->f_n(x)dx{\to <!--\; \to \; -->}_n\int <!--\; \int \; -->g(x)\cdot <!--\; \cdot \; -->f(x)dx$
holds. Then it follows that $f_n\to <!--\; \to \; -->f$ almost everywhere.
Intuitively the statement looks false, but I can't find a counterexample. If it doesn't hold: changes the further assumption that the $f_n,f$ have to be continuous anything?
Kind regards

All examples of a dense and co-dense set I have seen are either of full Lebesgue measure or of measure zero. For instance, in restriction to the unit interval $\mathbb{I}=[0,,1]$, we could have respectively $\mathbb{I}\cap <!--\; \cap \; -->\mathbb{Q}$ or $\mathbb{I}\setminus <!--\; \setminus \; -->\mathbb{Q}$. What I am looking for is a dense and co-dense subset $A\subset <!--\; \subset \; -->\mathbb{I}$ such that
$\mathrm{m}<!--\; \; -->(A)=\mathrm{m}<!--\; \; -->(\mathbb{I}\setminus <!--\; \setminus \; -->A)=\frac{1}{2}.$
I have attempted this task sequentially by, ever more finely, nibbling holes out of subintervals of $\mathbb{I}$ and partially back-filling the previously created holes. It's easy to approach half measure at each step, but I can't see how to to get convergence.

Let $(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{A})$ be a measurable space, and let $\mathrm{\mu <!--\; \mu \; -->}:\mathcal{A}\to <!--\; \to \; -->[0,+\mathrm{\infty <!--\; \infty \; -->})$ be a monotone subadditive set function. Let moroever $A_1,A_2,\dots <!--\; \dots \; -->\in <!--\; \in \; -->\mathcal{A}$ be a sequence of measurable sets. It is easy to prove that $\mathrm{\mu <!--\; \mu \; -->}(A_n)\to <!--\; \to \; -->0\Rightarrow <!--\; \Rightarrow \; -->\mathrm{\mu <!--\; \mu \; -->}(A_n^c)\to <!--\; \to \; -->\mathrm{\mu <!--\; \mu \; -->}(\mathrm{\Omega <!--\; \Omega \; -->})$. On the contrary, I am not able neither to prove the opposite implication, namely $\mathrm{\mu <!--\; \mu \; -->}(A_n)\to <!--\; \to \; -->\mathrm{\mu <!--\; \mu \; -->}(\mathrm{\Omega <!--\; \Omega \; -->})\Rightarrow <!--\; \Rightarrow \; -->\mathrm{\mu <!--\; \mu \; -->}(A_n^c)\to <!--\; \to \; -->0$, nor to find an example violating it. Can someone find either one or the other?

A ring is closed under intersection. And delta-ring is closed under countable intersection. According to my understanding, if a family of sets is closed under intersection, than it should also be closed under countable intersection.
Let's say $R$ is a ring. To $R$ be a delta-ring, $A_n\cap <!--\; \cap \; -->A_{n+1}\cap <!--\; \cap \; -->A_{n+2}\cap <!--\; \cap \; -->...$ should also be in $R$. And, as a ring is closed under intersection, $A_n\cap <!--\; \cap \; -->A_{n+1}$ is in $R$, then $A_n\cap <!--\; \cap \; -->A_{n+1}\cap <!--\; \cap \; -->A_{n+2}$ is in $R$, and so on. So, any ring is a delta-ring.
What am I missing? I guess I don't properly understand what countable intersection is. Thank you for reading!

Three distinct positive integers are selected at random from 9 consecutive positive integers. What is the probability that their average is also an integer?
Is there any way to find it efficiently? I'd taken example of 1,2,3,4,5,6,7,8,9 then there are a lot of combinations to give a number divisible by three which give an integer average value. It's difficult to find the way :D
Thank you.

Setting We work on a filtered probability space with finite time horizon $T$. The filtration is assumed to be complete. Let $X$ be a stochastic process that satisfies a property (A) a.s. For example, if property (A) is being nonnegative at the time $T$, then $X$ satisfies $P[X_T\ge <!--\; \ge \; -->0]=1$.
Question I would like to obtain that $X$ (up to indistinguishability) satisfies property (A) for each $\mathrm{\omega <!--\; \omega \; -->}$.
My attempt Define a process $Y$ to be equal to 0 on the event $\{X_T<0\}$ and define it to be equal to $X$ on $\{X_T\ge <!--\; \ge \; -->0\}$. Then $Y_T\ge <!--\; \ge \; -->0$. Is it then correct that $X,Y$ are indistinguishable, i.e. $P[X_t=Y_t,\mathrm{\forall <!--\; \forall \; -->}t\in <!--\; \in \; -->[0,T]]=1$? Is this approach possible whenever the event where the stochastic process does not satisfy the desired propert has probability 0?

In measure theory I encountered Egorov's theorem which states that if $(X,\mathcal{S},\mathrm{\mu <!--\; \mu \; -->})$ is a measure space such that $\mathrm{\mu <!--\; \mu \; -->}(X)<\mathrm{\infty <!--\; \infty \; -->}$ i.e. $\mathrm{\mu <!--\; \mu \; -->}$ is a finite measure.If $(f_n)$ be a sequence of measurable functions on $X$ converging pointwise to $f:X\to <!--\; \to \; -->\mathbb{R}$,then $f_n$ is almost uniformly convergent to $f$.
Now in the book the definition of almost uniform convergence is the following:
$f_n\to <!--\; \to \; -->f$ almost uniformly if for each $\mathrm{ϵ<!--\; ϵ\; -->}>0$,there exists $E_{\mathrm{ϵ<!--\; ϵ\; -->}}\subset <!--\; \subset \; -->X$ such that $\mathrm{\mu <!--\; \mu \; -->}(E_{\mathrm{ϵ<!--\; ϵ\; -->}})<\mathrm{ϵ<!--\; ϵ\; -->}$ and $f_n\to <!--\; \to \; -->f$ uniformly on $X-<!--\; -\; -->E_{\mathrm{ϵ<!--\; ϵ\; -->}}$.
Now in some books I have seen that almost everywhere means outside a measure 0 set.
So the definition of almost uniformly convergent should have been $f_n\to <!--\; \to \; -->f$ almost uniformly if $\mathrm{\exists <!--\; \exists \; -->}E\subset <!--\; \subset \; -->X$ such that $\mathrm{\mu <!--\; \mu \; -->}(E)=0$ and $f_n\to <!--\; \to \; -->f$ uniformly on $X-<!--\; -\; -->E$.
But unfortunately the definition is not so.In fact the latter condition is stronger.I want to know why the former is taken as a definition and what the problem with the latter one is.I want to understand where I am making mistake in understanding the word "almost".

One of the "Remarks" in our lecture notes says that if $\mathcal{G}$ is non-empty family of sets and ${\mathrm{\sigma <!--\; \sigma \; -->}}_{\mathcal{R}}(\mathcal{G})$ is the sigma ring generated by $\mathcal{G}$, then every set in ${\mathrm{\sigma <!--\; \sigma \; -->}}_{\mathcal{R}}(\mathcal{G})$ is a subset of countable union of sets in $\mathcal{G}$. Why is this remark true.

We are given an integrable function $f:[0,1]\to <!--\; \to \; -->\mathbb{R}$ and a set $A=\{(x,y)\in <!--\; \in \; -->{\mathbb{R}}^2|x,y\ge <!--\; \ge \; -->0,x+y\le <!--\; \le \; -->1\}$. I need to show that for $\mathrm{\alpha <!--\; \alpha \; -->},\mathrm{\beta <!--\; \beta \; -->}>0$ the following holds:
${\int <!--\; \int \; -->}_{A}f(x+y)x^{\mathrm{\alpha <!--\; \alpha \; -->}}{y}^{\mathrm{\beta <!--\; \beta \; -->}}d{\mathrm{\lambda <!--\; \lambda \; -->}}^2(x,y)={\int <!--\; \int \; -->}_0^1{y}^{\mathrm{\alpha <!--\; \alpha \; -->}}(1-<!--\; -\; -->y{)}^{\mathrm{\beta <!--\; \beta \; -->}}d\mathrm{\lambda <!--\; \lambda \; -->}(y){\int <!--\; \int \; -->}_0^{1}f(x)x^{\mathrm{\alpha <!--\; \alpha \; -->}+\mathrm{\beta <!--\; \beta \; -->}+1}d\mathrm{\lambda <!--\; \lambda \; -->}(x).$
We are provided with the hint to use the following map $\mathrm{\varphi <!--\; \varphi \; -->}:[0,1{]}^2\to <!--\; \to \; -->{\mathbb{R}}^2,\mathrm{\varphi <!--\; \varphi \; -->}(x,y)=(xy,x(1-<!--\; -\; -->y)).$. I do not see how to use the hint. It is clear that in addition to $f$ all other functions that one finds in the integrals on both sides are integrable on the set $A$ and their product as well. Thus, once one can figure out how to separate the variables, the claim will follow by using the Fubini theorem.
Can somebody provide some insight or a solution proposal? Thanks.