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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?

lmaosome
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1 Answers1

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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 $L_i(a_1,a_2,...,a_n)=a_i, i=1,2...,n$.

Kavi Rama Murthy
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  • oh, ok. that is indeed obvious (once you think a second about it). thank you – lmaosome Feb 12 '21 at 00:21
  • 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. – GEdgar Feb 12 '21 at 01:06