# Iterated Integral question and Fubini's theorem

$$\begin{array}{l}{\text { (3) Let } f \text { be an integrable function. Express the integral }} \\ {\qquad \int_{0}^{1} \int_{y}^{2 y} \int_{0}^{x+y} f(x, y, z)\, d z\, d x \,d y} \\ {\text { as a sum of iterated integrals in the order } d x d y d z \text { . }}\end{array}$$

For this to be in order $$dx\, dy\, dz$$, we need $$f$$ to be a continuous function and partial derivative need to exist but no condition is given as such so how to tackle such kind of function?

• No? If $f\in L^1(\mathbb{R}^3)$ (or $L^1([0,1]\times [0,2]\times [0,3])$ as seems to be your case), then the interchangability of integration order is part of the Fubini theorem. – WoolierThanThou Oct 10 '19 at 6:52
• @WoolierThanThou so what should be answer can you explain please? – maths student Oct 10 '19 at 7:22

The answer is $$\int_0^{3}\int_0^{1} \int_{\max \{z-y,y\}}^{2y} f(x,y,z)\,dx\,dy\,dz$$. Justification is by Fubini's Theoorem. [ Note that the inequalities $$0 are equivalent to the inequalities $$\max \{z-y,y\} and $$0].
• $x$ can on;y go up to $2y$. Of course, $2y <2$ but the limit of integration w.r..t $x$ has to go up to $2y$ not $2$. @mathsstudent – Kavi Rama Murthy Oct 10 '19 at 7:35
• @mathsstudent Once you have taken care of all the restrictions on $x$ and $y$ you have to integrate w.r.t. $z$ from its minimum value to its maximum value. [No other variable can appear in the final integral]. The maximum value of $z$ is $3$ because the maximum value of $y$ is $2$ and the maximum value of $x$ is $1$. – Kavi Rama Murthy Oct 10 '19 at 7:40