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I was reading this question here

Fubini's Theorem double integral with sin and $e^{-x}$ but I do not know why are we allowed to replace the integral with respect to the product measure $\mu$ with iterated integrals? could anyone explain this for me please?

Emptymind
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    Fubini says this can be done if $$ \int\limits_E \left| \frac y x e^{-x} \sin x , d \mu\right| <+\infty. \tag 1 $$ And we have $\left| \dfrac{\sin x} x \right| \le 1$ for all $x\in\mathbb R.$ So the integral in line $(1)$ above is $\displaystyle \le \int_0^\infty \sqrt x , e^{-x} , dx < +\infty. \qquad$ – Michael Hardy Feb 25 '20 at 04:17
  • @MichaelHardy why our integrand is measurable? – Emptymind Feb 25 '20 at 05:15
  • https://math.stackexchange.com/questions/3559331/why-are-we-allowed-to-replace-the-integral-with-respect-to-the-product-measure could you please look at this if you have time?@MichaelHardy – Emptymind Feb 25 '20 at 06:28

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We have that $E=\{(x,y)\in \mathbb R^2:0\le y\le \sqrt x\}.\ $

If we can show that that $\int_E \left|\frac{y}{x} e^{-x}\sin x\right|\ d(m\times m)$ is finite, then the result follows by Fubini's theorem.

To do this, we split $E$ into a union of the two sets

$E_1=\{(x,y):0\le y\le \sqrt x;\ 0 \le x\le 1\}$ and $E_2=\{(x,y):0\le y\le \sqrt x;\ x\ge 1\}$.

Then, on $E_1,\ \left|\frac{y}{x} e^{-x}\sin x\right|\le 2ye^{-x}$, and the integral of this over $E_1$ is finite.

On $E_2,\ \left|\frac{y}{x} e^{-x}\sin x\right|\le \sqrt xe^{-x}$, and the integral of this function over $E_2$ is also finite.

Matematleta
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  • how do you get your bounds for $E_{1}$ and $E_{2}$? – Emptymind Feb 25 '20 at 05:04
  • why our integrand is measurable? – Emptymind Feb 25 '20 at 05:14
  • In $E_{1}$ does not the bound should be function of $x$ only? – Emptymind Feb 25 '20 at 05:32
  • My statement for Fubini is this: My statement for Fubini theorem is:

    {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 $X \times Y.$ if $f$ is a real-valued and if $\psi^(x) = \int_{Y}|f_{x}|d\lambda$ and if $\int_{X} \phi^ d\mu < \infty$ then $f\in L^1(\mu \times \lambda).$ could you please show me how you tried to fullfil the requirements as I am confused a little bit ?

     – Emptymind Feb 25 '20 at 05:55
  • https://math.stackexchange.com/questions/3559331/why-are-we-allowed-to-replace-the-integral-with-respect-to-the-product-measure could you please look at this if you have time? – Emptymind Feb 25 '20 at 06:28
  • Are E_1 and E_2 the x sections and y sections of E? – Emptymind Feb 25 '20 at 13:44
  • @Emptymind If you draw the picture, you'll see that $E_1$ and $E_2$ are just the regions to the left and right, resp, of the vertical line from the point $(1,1)$ on the graph of $y=\sqrt x$ to the $x$-axis. measurablility of the integrands follows from continuity. If you want to use your definition of the statement of Fubini. apply it directly to the sections of $E$ and you will see that your integrals as functions of $x$ and $y$,resp are finite. On the other hand, I think it is easier to use the more standard statement of Fubini. – Matematleta Feb 25 '20 at 16:38