Proof that $\|F_B\|\leq\|F\|$.

Background Information

Let $(X,\mathscr{A},\mu)$ be a measure space, let $p$ satisfy $1\leq p<+\infty$. Let $F$ be an arbitrary element of $(L^p(X,\mathscr{A},\mu))^*$, so $F$ is a continuous linear functional on $L^p(X,\mathscr{A},\mu)$.

Suppose that $B$ belongs to $\mathscr{A}$. Let $\mathscr{A}_B$ be the $\sigma$-algebra on $B$ consisting of those subsets of $B$ that belong to $\mathscr{A}$. Let $\mu_B$ be the restriction of $\mu$ to $\mathscr{A}_B$. If $f$ is a real- or complex-valued function on $B$, then we will denote by $f'$ the function on $X$ that agrees with $f$ on $B$ and vanishes outside $B$. The formula $F_B(\langle f\rangle)=F(\langle f'\rangle)$ defines a linear functional $F_B$ on $L^p(B,\mathscr{A}_B,\mu_B)$. Show that this functional satisfies $\|F_B\|\leq\|F\|$.

What I Have Tried So Far

Let $\langle f\rangle\in L^p(B,\mathscr{A}_B,\mu_B)$, so that $f\in\mathscr{L}^p(B,\mathscr{A}_B,\mu_B)$, which implies $|f|^p$ is integrable (on $(B,\mathscr{A}_B,\mu_B)$), that is $\int|f|^pd\mu_B<+\infty$. But $\int|f|^pd\mu_B=\int|f'|^pd\mu$. Therefore,\begin{align*} F_B(\langle f\rangle) &= F(\langle f'\rangle)\\&\leq \|F\|\cdot\|\langle f'\rangle\|_p\\&= \|F\|\cdot\|f'\|_p\\&= \|F\|\cdot\left(\int|f'|^pd\mu\right)^{\frac{1}{p}}\\&= \|F\|\cdot\left(\int|f|^pd\mu_B\right)^{\frac{1}{p}}\\&= \|F\|\cdot\|f\|_{p,B}\\&= \|F\|\cdot\|\langle f\rangle\|_{p,B},\end{align*}where $\|\cdot\|_{p,B}$ is the norm defined on $\mathscr{L}^p(B,\mathscr{A}_b,\mu_B)$ or on $L^p(B,\mathscr{A}_B,\mu_B)$. Thus $\|F\|$ is a bound for the linear operator $F_B$. So $\|F_B\|$ as the infimum of the set of bounds for must be less than or equal to $\|F\|$.

My Question

1. I am not sure whether it is legit to say $\int|f|^pd\mu_B=\int|f'|^pd\mu$, because they are defined on different spaces.

2. Moreover, I am not sure whether it is legit to define a $\|\cdot\|_{p,B}$ and then use it as how I did in the equations.

I would really appreciate it if someone would help me with these questions!