Prove that the norm of an integrable vector function is integrable This result seems basic but I couldn't find a proof anywhere. Suppose $\mathbf{f}:[a,b]\to\mathbb R^n$ is Riemann integrable. Let $\|\cdot\|$ be a norm on $\mathbb R^n$. I want to show that $\|\mathbf f(x)\|$ is Riemann integrable as a function $[a,b]\to\mathbb R$. Is this result true? How can I prove it? Thanks in advance!
I know how to prove this result for specific norms like $\|\cdot\|_1$ and $\|\cdot\|_2$, but I don't know how to prove it for a general norm. btw I'm using the defintion that $\mathbf{f}:[a,b]\to\mathbb R^n$ is Riemann integrable if each of its components $f_i$ are Riemann integrable. Then the integral would be $\int\mathbf{f}(x)dx=(\int f_1(x)dx,\cdots,\int f_n(x)dx)$.
 A: Here is a formal proof of the result using the method Max has suggested:
We make use of the result $f: [a, b] \to \mathbb R$ is Riemann integrable if and only if $U_{\mathcal{D}_m}f -L_{\mathcal{D}_m}f \to 0$ as $m\to\infty$, where $\mathcal{D}_m$ is the dissection $a = x_0 < x_1 < \cdots < x_m = b$ given by $x_i = a + \frac{i(b - a)}{m}$ for each $i$. Write $I_k=[x_{k-1},x_k]$. To show that $\|\mathbf{f}(x)\|$ is integrable, note that 
\begin{align*}
\sup_{I_k}\|\mathbf f(x)\|-\inf_{I_k}\|\mathbf f(x)\|&=\sup_{x,y\in I_k}(\|\mathbf f(x)\|-\|\mathbf f(y)\|)\leq\sup_{x,y\in I_k}\|\mathbf f(x)-\mathbf f(y)\|\\
&\leq C\sup_{x,y\in I_k}\|\mathbf f(x)-\mathbf f(y)\|_1=C\sup_{x,y\in I_k}\left(\sum^n_{i=1}|f_i(x)-f_i(y)|\right)\\
&\leq C\sum^n_{i=1}\sup_{x,y\in I_k}|f_i(x)-f_i(y)|=C\sum^n_{i=1}\left(\sup_{I_k}f_i(x)-\inf_{I_k}f_i(y)\right)
\end{align*}
where we have use $\|\mathbf x\|-\|\mathbf y\|\leq\|\mathbf x-\mathbf y\|$ (triangle inequality) and $\|\mathbf x\|\leq C\|\mathbf x\|_1$ (all norms on $\mathbb R^n$ being equivalent). So
\begin{align*}
&U_{\mathcal{D}_m}\|\mathbf f\| -L_{\mathcal{D}_m}\|\mathbf f\|=\sum^m_{k=1}|I_k|\left(\sup_{I_k}\|\mathbf f(x)\|-\inf_{I_k}\|\mathbf f(x)\|\right)\\
&\leq C\sum^n_{i=1}\left(\sum^m_{k=1}|I_k|\left(\sup_{I_k}f_i(x)-\inf_{I_k}f_i(y)\right)\right)=C\sum^n_{i=1}(U_{\mathcal{D}_m}f_i -L_{\mathcal{D}_m}f_i)\to 0
\end{align*}
as $m\to\infty$. Hence $\|\mathbf f(x)\|$ is integrable.
A: For $i=1,..,n$ and $n\in\mathbb{N}$ let $f_i$ be Riemann-integrable i.e. $\int_{\mathrm{Dom}(f)}\left|f_i\right|\mathrm{d}x<\infty$ be finite and let $\mathrm{Dom}(f)$ be the domain of $f$. Thus with $\left\|v\right\|_1:=\sum_{i=1,...,n}\left|v\right|$ for $v\in\mathbb{R}^n$ it holds
\begin{equation}
\infty>\sum_{i=1,...,n}\left|f_i\right|=\int_{\mathrm{Dom}(f)}\sum_{i=1,...,n}\left|f_i(x)\right|\mathrm{d}x=\int_{\mathrm{Dom}(f)}\left\|f(x)\right\|_1\mathrm{d}x.
\end{equation}
Let $\left\|\cdot\right\|_0$ be any norm on $\mathbb{R}$. Since all norms on $\mathbb{R}^n$ are equivalent (thus also $\left\|\cdot\right\|_0$ and $\left\|\cdot\right\|_1$) we habe a univesal constant $C$ such that $\forall v\in\mathbb{R}^n : C\left\|v\right\|_0\leq \left\|v\right\|_1$. Pluging that inequality into the above inequality we get
\begin{equation}
\infty>\int_{\mathrm{Dom}(f)}\left\|f(x)\right\|_1\mathrm{d}x\geq \int_{\mathrm{Dom}(f)} C\left\|f(x)\right\|_0\mathrm{d}x.
\end{equation}
Hence by linearity of the integral operator $\int_{\mathrm{Dom}(f)}\left\|f(c)\right\|_0\mathrm{d}x$ must be finite.
EDIT: if you are lackong some theorems and need a proof for the riemann integral as limit of riemann sums, just do the same calculation with the riemann sums instead of the integral.
