# Metric Given by Riemann integral

Let $L_p([a,b];\mathbb{R}) = \{f:[a,b]\to\mathbb{R} | f$ is Riemann-integrable $\}$. I want to prove that $$d_p(f,g) = \left(\int_a^b |f(x)-g(x)|^p dx\right)^{1/p}$$ defines a metric in $L_p([a,b],\mathbb{R}).$

Given $f,g$ integrable I easily see that $d_p(f,g)$ exists and that $d_p(f,f)=0$ and $d_p(f,g) = d_p(g,f)$. I also proved the triangle inequality by proving the Hölder inequality for positive real functions. I'm just stuck in the condition $f\neq g \to d_p(f,g)>0$. For simplicity I'm assuming $f \neq 0$ and trying to prove $\int |f|^p \neq 0$.

I know that $f\ge 0$ implies $\int f \ge 0$ but I don't see how to get the strict inequality. Any help or hint will be appreciated.

• Actually the space should be altered a bit, since $f =0$ almost everywhere on $[a,b]$ would deduce that $\int_a^b |f|^p = 0$. – xbh Sep 5 '18 at 9:24
• Will continuos functions work? – AnalyticHarmony Sep 5 '18 at 13:03
• Yes, that would avoid the possibility that $\int |f - g|^p = 0$ but $f \neq g$, since $\int |f|^p = 0$ and $f$ be continuous would deduce that $f \equiv 0$. This claim could be proved by contradiction. – xbh Sep 5 '18 at 13:08
• So the usual notation $L_p$ are used for spaces of continuous functions? – AnalyticHarmony Sep 5 '18 at 13:09
• Not exactly. Actually $L$ stands for "Lebesgue" integration. Generally $R$ would be used for Riemann integrals. For continuous functions, try $C$, or $C_p$ to indicate the metric. – xbh Sep 5 '18 at 13:12

You can't prove it because it is not true. If$$f(x)=\begin{cases}1&\text{ if }x=a\\0&\text{ otherwise,}\end{cases}$$then $d_p(f,0)=0$, but $f\neq0$. The function $d_p$ is a pseudometric, but not a metric.
• @AnalyticHarmony Yes, if you restrict $d_p$ to the space $\mathcal{C}([a,b])$, then it becomes a metric. – José Carlos Santos Sep 5 '18 at 16:31