Changing limits when switching order of integration.

Question

Consider the following Lebesgue integral: $$ \int_{0}^{\infty} \int_{|f|>y} g(y) \: |f(x)| \: \mathrm{d}x \: \mathrm{d}y , $$ where g is a non-negative measurable function defined on $ (0,\infty) $ and $ f $ is a measurable function defined on $ \mathbb{R}^n $. If I am correct, the Tonelli's version of the Fubini theorem allows us to switch the order of integration. Question: What will be the new limits of integration?

My attempt: \begin{align} \{(x,y) : |f(x)|>y, \: y \in (0,\infty)\} &= \{(x,y) : x \in \mathbb{R}^n, \: |f(x)|>y, \: y \in (0,\infty)\} \\ &= \{(x,y) : x \in \mathbb{R}^n, \: |f(x)|>y>0\} . \end{align} Hence, the above double integral equals $$ \int_{\mathbb{R}^n} \int_{0}^{|f|} g(y) \: |f(x)| \: \mathrm{d}y \: \mathrm{d}x . $$ While I have no doubt that the the two integrals are indeed equal, I'm wondering whether the set-theoretic justification I gave is sufficient, or whether I missed some subtlety?

Many thanks in advance! :)

Answer 1

Your general thoughts are right but you made some (minor) mistakes or should consider points you might not realized before:

1.) you missed to switch "dx" and "dy" at the end, so the correct term is $$\int_{\mathbb{R}^n} \int_{0}^{|f|} g(y) \: |f(x)| \: \mathrm{d}y \: \mathrm{d}x$$

2.) You should recognize that actually your domain of the outer integral should be $$\Bbb R^n\setminus\{|f| \not=0 \}$$.

Because the integrand contains $|f|$ as a factor and so equals $0$ if $|f|=0$ this doesn't matter in your case but if the integrand would be e.g. $$g(x)\left(|f(x)| + 1\right)$$ it could!

Just be aware of that....

Answer 2

If the set $E$ is defined as

$$ E = \{ (x, y) : |f(x)| > y \} $$

then the function $(x, y) \mapsto g(y)|f(x)| \mathbf{1}_E(x, y)$ is non-negative and measurable on $\mathbb{R}^n \times (0,\infty)$, and hence by the Tonelli's theorem

\begin{align*} \int_{0}^{\infty} \int_{|f(x)|>y} g(y)|f(x)| \, dxdy &= \int_{0}^{\infty} \int_{\mathbb{R}^n} g(y)|f(x)|\mathbf{1}_E(x, y) \, dxdy \\ &= \int_{\mathbb{R}^n} \int_{0}^{\infty} g(y)|f(x)|\mathbf{1}_E(x, y) \, dydx \\ &= \int_{\mathbb{R}^n} \int_{0}^{|f(x)|} g(y)|f(x)| \, dydx \end{align*}

That being said, the only subtlety you have to deal with is the measurability of the set $E$, but this is rather obvious from the measurability of the function $(x, y) \mapsto f(x) - y$.