Cauchy-Formula for Repeated Lebesgue-Integration

Recently, I came across the following statements. They were annotated as consequences of Fubini's Theorem but neither proof nor reference were given.

1. Let $f:[a,b]\times [a,b]\to\mathbb{R}$ be Lebesgue integrable. Then it holds: $$\int_a^b\int_a^y f(x,y)\,\mathrm{d}x\,\mathrm{d}y =\int_a^b\int_x^b f(x,y)\,\mathrm{d}y\,\mathrm{d}x\text{.}$$
2. Let $f:[a,b]\to\mathbb{R}$ be Lebesgue integrable and $x\in[a,b]$. Then it holds: $$\int_a^x\int_a^{x_1}\cdots\int_a^{x_{n-1}} f(x_n)\,\mathrm{d}x_n\cdots\,\mathrm{d}x_2\,\mathrm{d}x_1 =\int_a^x \frac{(x-t)^{n-1}}{(n-1)!}f(t)\,\mathrm{d}t\text{.}$$

Proof:

1. Let $$g:[a,b]\times [a,b]\ni(x,y) \mapsto\left\{\begin{array}{ll} 1 & \text{, }x\le y \\ 0 & \text{, otherwise}\end{array}\right\}\in\mathbb{R}\text{.}$$ Then $g$ is measurable since $g^{-1}(S)$ is open or closed for all $S\subseteq\mathbb{R}$. Since $g$ is non-negative, it is clearly Lebesgue integrable. Consequently, $h:=f g$ is Lebesgue integrable. By Fubini's Theorem it follows that $$\int_{[a,b]\times [a,b]} h(x,y)\,\mathrm{d}(x,y) =\int_a^b\int_a^b h(x,y)\,\mathrm{d}x\,\mathrm{d}y =\int_a^b\int_a^b h(x,y)\,\mathrm{d}y\,\mathrm{d}x\text{.}$$ Finally, $$\int_a^b\int_a^b h(x,y)\,\mathrm{d}x\,\mathrm{d}y =\int_a^b\int_a^b f(x,y) g(x,y)\,\mathrm{d}x\,\mathrm{d}y =\int_a^b\int_a^y f(x,y)\,\mathrm{d}x\,\mathrm{d}y$$ and $$\int_a^b\int_a^b h(x,y)\,\mathrm{d}y\,\mathrm{d}x =\int_a^b\int_a^b f(x,y) g(x,y)\,\mathrm{d}y\,\mathrm{d}x =\int_a^b\int_x^b f(x,y)\,\mathrm{d}y\,\mathrm{d}x$$ conclude the proof.
2. We proceed by induction with respect to $n$. For $n=1$ the statement is trivial. Assuming its correctness for $n=m-1\in\mathbb{N}$, we prove the statement for $n=m$. Application of the induction hypothesis yields $$\int_a^x\int_a^{x_1}\cdots\int_a^{x_{m-1}} f(x_m)\,\mathrm{d}x_m\cdots\,\mathrm{d}x_2\,\mathrm{d}x_1 =\int_a^x\int_a^{x_1} \frac{(x_1-t)^{m-2}}{(m-2)!}f(t)\,\mathrm{d}t\,\mathrm{d}x_1\text{.}$$ We show that $$g:[a,x]\times [a,x]\ni(t,u) \mapsto\frac{(u-t)^{m-2}}{(m-2)!}f(t)\in\mathbb{R}$$ is Lebesgue integrable. First of all, $(t,u)\mapsto\frac{(u-t)^{m-2}}{(m-2)!}$ is Lebesgue integrable since it is continuous. Moreover, $(t,u)\mapsto f(t)$ is Lebesgue integrable as tensor product of Lebesgue integrable functions. Eventually, $g$ is Lebesgue integrable as product of Lebesgue integrable functions. Finally, 1. yields \begin{align} \int_a^x\int_a^{x_1}\cdots\int_a^{x_{m-1}} f(x_m)\,\mathrm{d}x_m\cdots\,\mathrm{d}x_2\,\mathrm{d}x_1 & =\int_a^x\int_t^x \frac{(x_1-t)^{m-2}}{(m-2)!}f(t)\,\mathrm{d}x_1\,\mathrm{d}t\\ & =\int_a^x\int_t^x \frac{(x_1-t)^{m-2}}{(m-2)!}\,\mathrm{d}x_1\,f(t)\,\mathrm{d}t\\ & =\int_a^x\frac{(x-t)^{m-1}}{(m-1)!}\,f(t)\,\mathrm{d}t. \end{align}

My questions:

• Is the proof correct and complete?
• Are the arguments unnecessarily complicated? Can the proof be simplified?
• How would you prove the statements?
• Do you know a reference of the statements in a textbook or scientific article?

Edit: Corrections included.

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The first proof is ok. – AD. Jan 7 '12 at 15:03
In the second statement you might have missed $dx$? Otherwise, it does not have much to do with part 1. as I see it...? BTW the verb is "we prove" not "we proof" - proof is a subject (a thing). – AD. Jan 7 '12 at 15:06
If you want every necessary statement and step actually addressed, the proofs don't seem long; and it's probably the most natural way of showing this using fubini. – gnometorule Jan 7 '12 at 15:22
Thank you for your answers. Indeed, in the second statement the integrand has to be $f(x_n)$ and not $f(x)$ as I had mistakenly written before. Comments to my embarrassing English are appreciated, too. The original question is updated. – precarious Jan 7 '12 at 15:35