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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 $\int\mathbf{f}(x)dx=(\int f_1(x)dx,\cdots,\int f_n(x)dx)$.

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    hint (actually more than a hint): on $\mathbb{R}^n$ all norms are equivalent. Plug in that inequality under the integral. Start with the sum of absolutes of the components (which coresponds the 1-norm inside the integral by linearity) – 2017-01-06
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    Sorry, could you be more specific? So we have $a\|\mathbf f(x)\|_1\leq \|\mathbf f(x)\|\leq b\|\mathbf f(x)\|_1$ and we know $\|\mathbf f(x)\|_1=\sum_i |f_i(x)|$ is integrable. How does this help? – 2017-01-06
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    i will write an answer later this evening. – 2017-01-06
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    you could add your edit to your answer instead. (someone rejected your edit on my answer, it wasn't me.) – 2017-01-07

2 Answers 2

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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.

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    But how do you know $\int_{\text{Dom}(f)}\|f(x)\|_0dx$ is Riemann integrable? Like the inf of the upper sum equals the sup of the lower sum? Like the limit of riemann sums might not exist – 2017-01-06
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    by the constant of the norm-equalency you can bound each summand of a riemann sum – 2017-01-06
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    I don't understand. It seems like you're claiming that if $f,g$ are two functions such that $C |f(x)|\leq |g(x)|$ for all $x$ and $\int |g(x)| dx$ exist, then $\int |f(x)| dx$ exist. But this is false I think, eg. $g$ be constant function 1, and $f$ be the function that is 1 when $x\in\mathbb{Q}$ and $0.5$ when $x\in\mathbb{R}\setminus\mathbb{Q}$. – 2017-01-06
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    both limits (upper and lower) exist and are independant of the refinement. thus their difference tends to zero. look at each summand of that difference and apply the norm inequality, you end up with something like $\frac{1}{C}\cdot\text{ sequence tending to zero}$ proving that upper and lower darboux sequence tend to the same value. (wich must be finite since you can estimate both upper darboux sums with the original darboux sums by norm equalency again) – 2017-01-06
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    Sorry I can't see how this works, please write me an answer if possible! Thanks – 2017-01-06
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    Wlog let the interval [a,b]=[0,1]. We need to show that $\sum^N_{n=1}\frac{1}{N}(\sup_{x\in[\frac{n-1}{N},\frac{n}{N}]}\|f(x)\|_0-\inf_{x\in[\frac{n-1}{N},\frac{n}{N}]}\|f(x)\|_0)$ converges to 0 as $N\to\infty$. I don't see how I can do anything with this. – 2017-01-06
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    The norm equivalence inequality seems to be useless in this case. If f,g are two functions such that $a|g(x)|\leq |f(x)|\leq b|g(x)|$ for all x and $\int^b_a |g(x)|dx$ exist, then $\int^b_a |f(x)|dx$ need not exist. For example let g be constant function 1, and f be the function that is 1 when $x\in\mathbb{Q}$ and 0.5 when $x\in\mathbb{R}\setminus\mathbb{Q}$. This illustrate the equivalence inequality $a|g(x)|\leq |f(x)|\leq b|g(x)|$ on its own is not sufficent for the result, so you must have been using somthing more, what else have you been using? – 2017-01-06
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    For $x^0, x_0\in I_n$ such that $\left\|f(x_0)\right\|_0=\mathrm{inf}_{x\in I_n}\left\|f(x)\right\|_0, \left\|f(x^0)\right\|_0=\mathrm{sup}_{x\in I_n}\left\|f(x)\right\|_0$ you have $0\leq \left\|f(x^0)\right\|_0-\left\|f(x_0)\right\|_0 \stackrel{\text{triangle inequality}}{\leq} \left\|f(x^0)-f(x_0)\right\|_0$. can you get it from here? – 2017-01-07
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    (note that for riemann integral you better choose $x^0$ such that $\left|\left\|f(x^0)\right\|_0-\mathrm{sup}_{x\in I_n}\left\|f(x)\right\|_0\right|\leq \epsilon$ with e.g. $\epsilon=\frac{1}{N}$ or $\epsilon=\frac{1}{N^2}$ depending what you need for the calculations that i did not explicitely do. same for the infimum of course; just a tiny detail one might forget. for the general idea the above is easier to understand.) – 2017-01-07
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    I still don't see how this would work. Using what you said, the most that I can show is that $\sum^N_{n=1}\frac{1}{N}(\sup_{x\in[\frac{n-1}{N},\frac{n}{N}]}\|f(x)\|_0-\inf_{x\in[\frac{n-1}{N},\frac{n}{N}]}\|f(x)\|_0)$ is bounded for all $N$, but couldn't show convergence (to 0). – 2017-01-07
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    as i wrote in my comment, replace sup and inf by (approximating) function values ($\left\|f(x_0)\right\|_0, \left\|f(x^0)\right\|_0$) and then first use the triangle inequality, then the norm inequality and go on from there. the rest is straight forward. – 2017-01-07
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    So now don't even worry about whether those supremum/infimum are attained. Just assume they are all attained, then using your notation $0\leq \left\|f(x^0)\right\|_0-\left\|f(x_0)\right\|_0 \leq \left\|f(x^0)-f(x_0)\right\|_0\leq C\left\|f(x^0)-f(x_0)\right\|_1\leq C\|f(x^0)\|_1+C\|f(x_0)\|_1$. – 2017-01-07
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    So we have $$\sum^N_{n=1}\frac{1}{N}(\sup_{x\in[\frac{n-1}{N},\frac{n}{N}]}\|f(x)\|_0-\inf_{x\in[\frac{n-1}{N},\frac{n}{N}]}\|f(x)\|_0)\leq C\sum^N_{n=1}\frac{1}{N}\|f(x^0)\|_1+C\sum^N_{n=1}\frac{1}{N}\|f(x_0)\|_1$$ We know $C\sum^N_{n=1}\frac{1}{N}\|f(x^0)\|_1+C\sum^N_{n=1}\frac{1}{N}\|f(x_0)\|_1\to 2C\int^b_a \|f(x)\|_1dx$. So this shows that it's bounded, but how do you show it converges to 0? – 2017-01-07
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    instead of the application of triangle inequality at the last inequality in the first of your 2 latest comments use the exact definition of the 1-norm instead – 2017-01-07
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    How would $\|f(x^0)-f(x_0)\|_1=\sum^n_{i=1}|f_i(x^0)-f_i(x_0)|$ help? – 2017-01-07
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    $\left|f_i(x^0)-f_i(x_0)\right|\leq\left|\sup_{x\in I_n}f_i(x)-\inf_{x\in I_n}f_i(x)\right|$ – 2017-01-07
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    even better: $... \leq \sup -\inf $ without $\left|\cdot \right|$. – 2017-01-07
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    I see! Seems to work, thanks! I'll write the formal proof without the assupmtion of attaining sup/inf later and see if I encounter any more problem – 2017-01-07
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    i think all what you get is an extra $+\frac{1}{N}$. – 2017-01-07
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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.