Here is the statement of Fubini's theorem we are using(my professor said that it is from the book of Saks but I searched the book but could not find it, So if anyone knows from which book is this statement I would appreciate this too much.)
{Let $(X, \mathcal{S}, \mu)$ and $(Y, \mathcal{T}, \lambda)$ be $\sigma$-finite measure spaces. And let $f$ be a $\mathcal{S} \times \mathcal{T}$ measurable function on $X \times Y.$
(a) If $0 \leq f \leq \infty$ and $\varphi(x) = \int_{Y}f_{x}d\lambda(y), \psi(y) = \int_{X}f^{y}d\mu(x)$ then $\varphi$ is $\mathcal{S}-$measurable and $\psi$ is $\mathcal{T}-$measurable
and $$\int_{X} \varphi d\mu = \int_{X \times Y} f d(\mu \times \lambda) = \int_{Y} \psi d\lambda.\quad \quad (**)$$
(b) If $f$ is real-valued and if $\varphi^{*}(x) = \int_{Y} |f_{x}| d\lambda $ and if $\int_{X} \varphi^{*} d\mu < \infty $ then $f \in L^{1}(\mu \times \lambda).$
(c) If $f \in L^{1}(\mu \times \lambda)$ then $f_{x} \in L^{1} (\lambda)$ for a.e. $x \in X $ and $f^{y} \in L^{1} (\mu)$ for a.e. $y \in Y , \varphi $ and $\psi$ are in $L^{1} (\mu)$ and $L^{1} (\lambda)$ respectivily and $(**)$ holds.
Here is the question and a solution that took $0$:
Apply Fubini's Theorem to calculate $$\int_{E} \frac{y}{x} e^{-x} \sin x d\mu$$ where $\mu$ is the product of Lebesgue measure on $\mathbb{R}$ with itself, and $E = \{(x,y) : 0 \leq y \leq \sqrt{x}\}.$}\
\textbf{Proof:}\
First. Proving that we are allowed to replace the integral with respect to the product measure $\mu$ with iterated integrals by checking Fubini theorem part(b,c) assumptions. \
Since, by the given we have that, our spaces are $(X, \mathcal{M}, m )$ and $(Y, \mathcal{M}, m )$ which are $\sigma-$ finite.and our function $f = \frac{y}{x} e^{-x} \sin x $ belongs to $\mathcal{M} \times \mathcal{M}$ where $\mathcal{M}$ is the Lebesgue $\sigma-$ algebra of subsets of $X$ and $Y$ respectively. Our function $f$ should be in $\mathcal{M} \times \mathcal{M}$ for the product measure to be defined and it is $\mathcal{M} \times \mathcal{M}-$measurable which can easily be proved as our function is continuous except at $0$ i.e on a set of measure zero. and our $f$ is real valued. \
Now, we will calculate the integral $\varphi^*(x) = \int_{Y}|f_x|dm$ where $|f_x|$ is the x-section of the function $f$ defined by $f(x,.) = f(x,y)$ and it is measurable by a theorem and it is defined on $E_x$ the x-section of $E$ where $E$ is any subset of $X\times Y$ and $x$ is any point in $X.$ So, $\varphi^*(x) = \int_{Y}|f_x|dm = \int_{0}^{\sqrt{x}} |\frac{y e^{-x}\sin x}{x}|dm(y) = \frac{e^{-x}}{x} |\sin x| \int_{0}^{\sqrt{x}} y dm(y) = \frac{e^{-x}}{x} |\sin x|\frac{x}{2} = \frac{e^{-x}}{2}|\sin x| \leq \frac{e^{-x}}{2}. $\
Now, calculating the integral $\int_{X} \frac{e^{-x}}{2}dm(x)$ if we found it bounded then sure the integral $\int_{X} \varphi^{*} d\mu $ will be $<\infty.$ \
So, $\int_{X} \frac{e^{-x}}{2}dm(x) = \int_{0}^{\infty} \frac{e^{-x}}{2}dx = \frac{1}{2},$ it was improper integral so we took (after integrating) $\lim_{t\rightarrow \infty} \frac{-1}{2}(e^{-t} -1).$ So we have that $\int_{X} \varphi^{*} d\mu <\infty.$Then our function $f \in L^1(m \times m)$ by Fubini theorem part (b). Now by Fubini theorem part (c) we have that $f_x \in L^1(m)$ for a.e. $x \in X$ and $f^y \in L^{1}(m) $ for a.e.$y \in Y$ and $\varphi, \psi$ are in $L^1(m)$ both and $$\int_{X}\varphi dm(x) = \int_{X\times Y} f d(m\times m) = \int_{y}\psi dm(y)$$ where $\varphi(x) = m (\varphi_{x}) = \int_{Y} f(x,y)dm(y) $ and $\psi(y) = m(\varphi^y)= \int_{X}f(x,y)dm(x).$ \
And now, we are allowed to replace the integral with respect to the product measure $\mu$ with iterated integrals.\
\textbf{Calculation of the integral:}
$$\int_{E} \frac{y}{x} e^{-x} \sin x d\mu = \int_0^\infty \int_0^{\sqrt{x}} \frac{y}{x} e^{-x}\sin x\, dy\, dx.$$
Since
$y \le \sqrt{x},\ y^2 \le x$
so
$x \ge y^2.$
So, $$\begin{align}
I
&=\int_0^\infty \int_0^{\sqrt{x}} \frac{y}{x} e^{-x}\sin x\, dy\, dx\\[6pt]
&=\int_0^\infty \int_{y^2}^\infty \frac{y}{x} e^{-x}\sin x\, dx\, dy
\end{align}$$
Now,
$$I
=\int_{0}^{\infty} \int_{0}^{\sqrt{x}} \dfrac{y}{x} e^{-x}\sin x\, dy\, dx
=\int_{0}^{\infty} \dfrac{e^{-x}\sin x}{x} \int_{0}^{\sqrt{x}}y \, dy\, dx
=\int_{0}^{\infty} \dfrac{e^{-x}\sin x}{x}\dfrac{x}{2} dx
=\dfrac12\int_{0}^{\infty} e^{-x}\sin xdx.
$$
\
The last integral is calculated as follows: We will integration by parts twice in :\
1)$ \text{First time } (f=\sin(x), g'=e^{-x} \Leftrightarrow f'=\cos(x), g=-e^{-x}$):
$$\int e^{-x} \sin(x) dx = -e^{-x} \sin(x) – \int -e^{-x} \cos(x)$$
2)$ \text{Second time } (f=\cos(x), g'=-e^{-x} \Leftrightarrow f'=-\sin(x), g=e^{-x}$):
$$- e^{-x} \sin(x) – \int – e^{-x} \cos(x) = – e^{-x} \sin(x) – (e^{-x} \cos(x) – \int -e^{-x} \sin(x))$$
So now we get:
$$\int e^{-x} \sin(x) dx = – e^{-x} \sin(x) – (e^{-x} \cos(x) + \int e^{-x} \sin(x))$$
$$2 \int e^{-x} \sin(x) dx = – e^{-x} \sin(x) – e^{-x} \cos(x) $$
$$ \int e^{-x} \sin(x) dx = \frac{- e^{-x} \sin(x) – e^{-x} \cos(x)}{2} $$
Now you can evaluate the integral for the final answer:
$$ \int_{0}^{\infty} e^{-x} \sin(x) dx = \frac{1}{2}. $$
The justification why the mark was 0 is as follows:
1- What are $X$ and $Y$? it seems that you have taken $Y = [0, \sqrt{x}]$ which is impossible since $Y$ can not depend on $X.$
My questions are:
1-I do not really understand how the given statement of Fubini's theorem said that "$Y$ can not depend on $X.$", which part of the theorem said this?
2- Could anyone help me correcting the above solution so that I can know the correct answer to the question (maybe by telling me what are the correct $X$ and $Y$ that should be used)?
Best Answer
The spaces $\mathrm{X}$ and $\mathrm{Y}$ are given, as your teacher said, neither of them can depend in the other. If you believe that, say $\mathrm{Y}$ depends on the points of $\mathrm{X},$ then you simply do not understand at all what product measures are and need to go back to the very basics (the construction of integral). Having said that, I proceed with your exercise.
You should have written the integral as follows: $\displaystyle \int\limits_{\mathbf{R}^2} d\mu(x, y)\ \dfrac{y}{x} e^{-x} \sin(x) \mathbf{1}_{\{0 \leq x\}}\mathbf{1}_{\left\{0\leq y < \sqrt{x}\right\}}$ and then continued. In this way of writing it must be obvious that $\mathrm{X} = \mathrm{Y} = \mathbf{R};$ if you do not get this, you really need to go back and read what product measures are. Observe that $\mathrm{E}$ is, by definition, the set obtained as the intersection of the sets that I wrote in the indicator functions. (We are simply applying the definition of the integral over a set: for any measure $\nu$ on the measurable space $\mathrm{Z}$ and any measurable set $\mathrm{E}$ here, we have $\int\limits_\mathrm{E} d\nu\ \varphi = \int\limits_\mathrm{Z} d\nu\ \mathbf{1}_\mathrm{E} \varphi$ whenever either of this expressions make sense.)
To your exercise. First show that the given function is integrable. Bound $|\sin(x)| \leq 1.$ Next apply Tonelli's theorem which is item (a) in the theorem you wrote. (Remark: Tonelli's theorem is Fubini's theorem for non negative measurable functions that may or may not be integrable; Fubini is a corollary of Tonelli; Tonelli proved the stronger version after Fubini.) You can do $$ \begin{align*} \int\limits_{\mathbf{R}^2} d\mu(x, y)\ \dfrac{y}{x} e^{-x} \mathbf{1}_{\{0 \leq x\}}\mathbf{1}_{\left\{0\leq y < \sqrt{x}\right\}} &= \int\limits_{\mathbf{R}} dx\ \dfrac{e^{-x}}{x} \mathbf{1}_{\{0 \leq x\}} \int\limits_{\mathbf{R}} dy\ y \mathbf{1}_{\left\{0\leq y < \sqrt{x}\right\}} \\ &= \int\limits_{0}^\infty dx\ \dfrac{e^{-x}}{x} \int\limits_{0}^{\sqrt{x}} dy\ y, \end{align*} $$ or the other way around, $$ \begin{align*} \int\limits_{\mathbf{R}^2} d\mu(x, y)\ \dfrac{y}{x} e^{-x} \mathbf{1}_{\{0 \leq x\}}\mathbf{1}_{\left\{0\leq y < \sqrt{x}\right\}} &= \int\limits_{\mathbf{R}} dy\ y \mathbf{1}_{\{0 \leq y\}} \int\limits_{\mathbf{R}} dx\ \dfrac{e^{-x}}{x} \mathbf{1}_{\{0 \leq x\}}\mathbf{1}_{\left\{y < \sqrt{x}\right\}} \\ &= \int\limits_0^\infty dy\ y\int\limits_{y^2}^{\infty} dx\ \dfrac{e^{-x}}{x}. \end{align*} $$ Whichever way you choose, will show that the absolute value of the function you are trying to integrate is integrable, then you can apply Fubini's theorem (since $\sin(x)$ can be negative sometimes). Your actual calculations will hold. (These are ultimately not important. What is important is you being able to understand why you can decompose the integrals into reiterated integrals.)