Lebesgue Integral of coordinate function is integral of individual functions

Lang writes in his book Real and Functional Analysis that it is "obvious" for $$f:\Omega\rightarrow\mathbb R^n$$ with $$f(\omega) = (f_1(\omega), f_2(\omega), \dots, f_n(\omega))$$ the integral of $$f$$ is given by $$\int_\Omega f(\omega)\,\mu(\mathrm d\omega) = \begin{pmatrix}\int_\Omega f_1(\omega)\,\mu(\mathrm d\omega) \\ \int_\Omega f_2(\omega)\,\mu(\mathrm d\omega) \\ \vdots \\ \int_\Omega f_n(\omega)\,\mu(\mathrm d\omega) \end{pmatrix},$$ i.e. the integral of the coordinate function is the integral of the individual functions. I don't see why this result holds or is even obvious?

For any linear map $$L: \mathbb R^{n} \to \mathbb R$$ we have $$L(\int fd\mu)=\int (L\circ f) d\mu$$. This is an easy consequence of the definition of the integral. Apply this to the coordinate maps $$L_i(a_1,a_2,...,a_n)=a_i, i=1,2...,n$$.
• Plus the fact that (in finite-dimensional spaces) all linear functionals are continuous. This may be needed, depending on how $\int f\;d\mu$ is defined. Feb 12 at 1:06