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

• 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) – Max Jan 6 '17 at 17:16
• 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? – lamkingming Jan 6 '17 at 17:41
• i will write an answer later this evening. – Max Jan 6 '17 at 21:21

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 $$\infty>\sum_{i=1,...,n}\left|f_i\right|=\int_{\mathrm{Dom}(f)}\sum_{i=1,...,n}\left|f_i(x)\right|\mathrm{d}x=\int_{\mathrm{Dom}(f)}\left\|f(x)\right\|_1\mathrm{d}x.$$ 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 $$\infty>\int_{\mathrm{Dom}(f)}\left\|f(x)\right\|_1\mathrm{d}x\geq \int_{\mathrm{Dom}(f)} C\left\|f(x)\right\|_0\mathrm{d}x.$$ 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.

• 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 – lamkingming Jan 6 '17 at 22:03
• by the constant of the norm-equalency you can bound each summand of a riemann sum – Max Jan 6 '17 at 22:03
• 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 $x\in\mathbb{R}\setminus\mathbb{Q}$. – lamkingming Jan 6 '17 at 22:09
• 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) – Max Jan 6 '17 at 22:20
• Sorry I can't see how this works, please write me an answer if possible! Thanks – lamkingming Jan 6 '17 at 22:30

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.