$(\int f_1d\mu)^2+\cdots+(\int f_nd\mu)^2\leq(\int \sqrt{f_1^2+\cdots+f_n^2}d\mu)^2$

Here are a couple of strategies that work in general and make no use of any type of local integrability properties of the underlying measure ($\sigma$-finiteness or not).


Consider the space $L$ of functions $f:X\rightarrow\mathbb{R}^n$ which are integrable in each component and define $\|f\|^*=\int\|f\|_2\,d\mu$, where $\|\;\|_2$ is the Euclidean norm on $\mathbb{R}^n$. This is defines a norm on $L$ since $\|f\|^*\leq\sum^n_{k=1}\int|f|_j\,d\mu<\infty$. Also, $$ \int|\|f\|_2-\|g\|_2|\,d\mu\leq\int\|f-g\|_2\,d\mu=\|f-g\|^* $$

Consider $\mathcal{E}$ the collection of (integrable) simple functions on $(X,\mathscr{B},\mu)$ and define $$\mathbb{R}^n\otimes\mathcal{E}=\{\sum^m_{k=1}u_k\phi_k: u_k\in\mathbb{R}^n, \phi_k\in\mathcal{E}, m\in\mathbb{N}\}$$

This space will play the role of elementary functions in the construction of the real valued integral. It is easy to check that $\mathbb{R}^n\otimes\mathcal{E}$ is dense in $(L,\|\;\|^*)$; furthermore, any function in $\mathbb{R}^n\otimes\mathcal{E}$ can be expressed as $$ \Phi=\sum^{M}_{j=1}v_j\mathbb{1}_{A_j} $$ where $v_j\in \mathbb{R}^n$, $A_j\in\mathscr{B}$, $\mu(A_j)<\infty$, and $M\in\mathbb{N}$. Consider now the elementary integral $$\int\Big(\sum^m_{k=1}u_k\phi_k\Big):=\sum^m_{j=1}u_k\int\phi_k\,d\mu$$

Since $\Phi=\sum_{u\in\mathbb{R}^n}u\mathbb{1}_{\{\Phi=u\}}$ (notice that the sum over $\mathbb{R}^n$ is actually finite), $$ \int\Phi =\sum^m_{j=1}u_j\mu(A_j)=\sum_{u\in\mathbb{R}^n}u\int\mathbb{1}_{\{\Phi=u\}}\,d\mu\tag{1}\label{one} $$

which means that the elementary integral extended to $\mathbb{R}^n\otimes\mathcal{E}$ does not depend on any particular representation of $\Phi$. Now $$ \Big\|\int\Phi\Big\|_2\leq\sum_{u\in\mathbb{R}^n}\|u\|_2\int\mathbb{1}_{\{\Phi=u\}}\,d\mu=\int\Big(\sum_{u\in\mathbb{R}^n}\|u\|_2\mathbb{1}_{\{\Phi=u\}}\Big)\,d\mu=\int\|\Phi\|_2\,d\mu=\|\Phi\|^*\tag{2}\label{two} $$ $\eqref{two}$ is the inequality you are looking for but only for functions in $\mathbb{R}^n\otimes\mathcal{E}$. For all functions in $L$ one can use some density arguments.


Comments:

  1. Notice that $\|\;\|_2$ can be replaced by $\|\;\|_p$ ($p\geq1$).

  2. Your problem is an example of an integral defined on vector--valued functions.

  3. The arguments used, with some technical additions (Daniell integration, and measurability issues) can be used to construct Bochner's integral where $\mathbb{R}^n$ is replaced by a Banach space.


Another, much simpler solution may be obtained by applying linear functionals to the vector $\int f=\sum^n_{j=1}e_j\int f_j\,d\mu$ where $e_1,\ldots,e_n$ is the standard basis of $\mathbb{R}^n$. As above, w $\|\,\|_p$ is $p$-norm in $\mathbb{R}^n$. We use the fact that $(\mathbb{R}^n,\|;\|_p)$ and $(\mathbb{R}^n,\|\,\|_q)$ are dual to each other when $\tfrac1p+\tfrac1q=1$.

If $\Lambda:\mathbb{R}^n\rightarrow\mathbb{}$ is linear, then $\Lambda x =x\cdot u$ for some unique $u\in\mathbb{R}$. Thus

\begin{aligned} \Lambda \Big(\int f\Big) &= u\cdot\Big(\int f\Big)=\sum^n_{j=1}u_j\int f_j\,d\mu =\int u\cdot f\,d\mu \end{aligned} and so, by Hölder's inequality (in $\mathbb{R}^n$) \begin{aligned} \left|\Lambda \Big(\int f\Big)\right|&\leq\int|u\cdot f|\,d\mu\\ &\leq\int\|u\|_q\|f\|_p\,d\mu=\|u\|_q\int\|f\|_p\,d\mu \end{aligned} The result than follows by taking $\sup$ over all linear functionals $\Lambda$ with functional norm $\|\Lambda\|:=\sup_{\|x\|_p=1}|\Lambda x|\leq1$, or equivalently, by taking $\sup$ over all vectors $u\in\mathbb{R}^n$ with $\|u\|_q=1$. Thus

$$\left\|\int f\right\|_p \leq \int\|f\|_p\,d\mu$$