Let $f$ be integrable over $E$. Define $\hat{f}$ to be the extension of $f$ to all of $\mathbb{R}$ obtained by setting $\hat{f}$$\equiv 0$ outside of $E$. Show that $\hat{f}$ is integrable over $\mathbb{R}$ and $\int_E f$=$\int_{\mathbb{R}} \hat{f}$.
This is a problem from Royden 4th edition. I am unsure how to proof this problem. We know $f$ is integrable over $E$. And we have thm. $26$ in the book that has the condition $h\equiv0$. However, I think I may need to use the Vitali's convergence theorem instead? A complete proof or concrete steps are appreciated.
First, let's show that $\hat{f}$ is Lebesgue integrable on $\mathbb{R}$. Consider the set $E_{\alpha} = \{ x\in \mathbb{R}: \hat{f}(x) \geq \alpha\}$ for $\alpha \in \mathbb{R}$. We have two cases
$$\alpha \leq 0: E_{\alpha} = \{ x\in \mathbb{R}: \hat{f}(x) \geq \alpha\} = (\mathbb{R}\setminus E) \cup \{x\in E: f(x) \geq \alpha\}$$
$$\alpha > 0: E_{\alpha} = \{ x\in \mathbb{R}: \hat{f}(x) \geq \alpha\} = \{x\in E: f(x) \geq \alpha\}$$
So, it's Lebesgue measurable for any $\alpha \in \mathbb{R}$. This shows that $\hat{f}$ is integrable on $\mathbb{R}$. A simple calculation shows that
$$\int_\mathbb{R} \hat{f} = \int_{\mathbb{R}\setminus E} \hat{f} + \int_E \hat{f} $$
The first term in the RHS is zero. Also, the second term is equal to $\int_E f$. Therefore $$\int_\mathbb{R} \hat{f} = \int_E f $$