Problem. Let $E$ be a domain in the plane bounded by the continuous curves $y = \phi(x)$ and $y = \psi(x)$ for $x \in I = [a,b]$, where $\phi(x), \psi(x)$. Prove that if $f$ is a Borel measurable, integrable function defined on $E$, then $$\int_{E}f = \int_{I}\int_{[\phi(x),\psi(x)]}f(x,y)dydx.$$
My attempt. Use the Fubini's Theorem that says $$\int_{I}\int_{[\phi(x),\psi(x)]}f(x,y)dydx = \int_{I \times [\phi(x),\psi(x)]}f(x,y)d(y \times x)$$ where $x \times y$ represents the product measure. But I could not relate $\int_{E}f$ with $\int_{I \times [\phi(x),\psi(x)]}f$. I believe that this is the main point of the question. Can someone help me?
The idea is to apply Fubini's theorem in $\mathbb R^2.$ Recall that for any $g$ Borel measurable and integrable on $\mathbb R^2,$ we have, $$ \int_{\mathbb R^2} g \ = \int_{\mathbb R}\int_{\mathbb R} g(x,y)\,\mathrm{d}x\,\mathrm{d}y. $$ The key trick is to embed the information about $E$ into this function $g.$ So we set, $$ g = \chi_E f. $$ Where $\chi_E$ is the characteristic function of $E \subset \mathbb R^2.$ By our assumptions $g$ is indeed Borel measurable and integrable, such that, $$ \int_{\mathbb R^2} \chi_Ef = \int_E f = \int_{\mathbb R}\int_{\mathbb R} \chi_E(x,y)f(x,y)\,\mathrm{d}x\,\mathrm{d}y. $$ Now $(x,y) \in E$ if and only if $x \in I$ and $\varphi(x) \leq y \leq \psi(x).$ So we conclude that, $$ \int_{\mathbb R}\int_{\mathbb R} \chi_E(x,y)f(x,y)\,\mathrm{d}x\,\mathrm{d}y = \int_I \int_{\psi(x)}^{\varphi(x)} f(x,y)\,\mathrm{d}x\,\mathrm{d}y, $$ from which the result follows.