Suppose $g : [−1,1] \to\mathbb R$ is Riemann integrable on $[−1,1], g(x) ≥ 0$ for all $x ∈ [−1,1]$, and $g(0) > 0$. Does it follow that $\int_{-1}^{1} g > 0$?
I tried proving that it is true by the definition of Riemann integration. Let $P={x_0,x_1,...,x_n}$ be a partition of $[-1,1]$. So, the Upper sum $U(g,P)=\sum M_i\Delta x_i$ where $M_i= Sup \{g(x): x\in [x_{i-1},x_i]\}$. As $g(0)>0$ and $g(x)$ is non-negative in this interval, we may write $U(g,P)>0$. But my doubt is, this implies that the upper integral $U(g)=Inf \{U(g,P), P$ is a partition of $[-1,1]\}\geq 0$. How do we get the strict inequality?