I have a question: Let $f:[a,b]\to\Bbb R$ be continuous and $\forall x,~f(x)\geq 0$. Suppose $f(x_0)>0$ for some $x_0\in[a,b]$, I was asked to prove that $\int_a^bf(x)>0$. I want someone help me to check if my proof is wrong or having a weak point. And I even want to know whether my proof is too silly or redundant, or could be better.
My attempt: I have already proved the fact that $\int_a^bf(x)\ge 0$ (skipped). So I only need to show that $\int_a^bf(x)>0$.
Suppose $\int_a^bf(x)=0$.
Since $f$ is continous at $x_0$, by the definition of continuity, there exists $δ>0$ such that $∀x\in[a,b],~|x-x_0|<δ\Rightarrow |f(x)-f(x_0)|<\frac{f(x_0)}{2}$.
Choose $δ'=\frac{δ}{2}$, then $∀x\in[a,b],~|x-x_0|≤δ'\Rightarrow |f(x)-f(x_0)|<\frac{f(x_0)}{2}$.
Since $\begin{alignedat}[t]{2} &|f(x)-f(x_0)|<\frac{f(x_0)}{2}\\ \Rightarrow ~&-\frac{f(x_0)}{2}<f(x)-f(x_0)<\frac{f(x_0)}{2}\\ \Rightarrow ~&-\frac{f(x_0)}{2}+f(x_0)<f(x)\\ \Rightarrow ~&\frac{f(x_0)}{2}<f(x)\\ \end{alignedat}$,
we have $∀x\in[a,b],~|x-x_0|≤δ'\Rightarrow f(x)>\frac{f(x_0)}{2}$.
Since $f(x_0)δ'>0$, apply this positive number to the definition of $\int_a^bf(x)=0$, we deduce that there exists $\eta>0$ such that:
$∀P:\text{partition of }[a,b],~∀T:\text{sample set of }P,~\|P\|<\eta\Rightarrow |R(P,T)-0|=|R(P,T)|<f(x_0)δ'$
Let $P$ be any partition so that $\|P\|<\eta$ and $x_0-δ',~x_0+δ'\in P$. Let $T$ be any sample set of $P$.
Let $P'\subseteq P$ and $P'=\{x_0-δ'=y_0,y_1,~y_2,~y_3,~\cdots,~x_0+δ'=y_n\}$.
Then $\begin{alignedat}[t]{2}R(P,T)&≥R(P',T)~~~(\text{Because }∀x,~f(x)≥0)\\ &=\sum_{i=0}^{n}f(\xi_i)\Delta y_i>\sum_{i=0}^{n}\frac{f(x_0)}{2}\Delta y_i=\frac{f(x_0)}{2}\sum_{i=0}^{n}\Delta y_i=\frac{f(x_0)}{2}2δ'=f(x_0)δ'\end{alignedat}$.
On the other hand, by the definition of integral, $|R(P,T)|<f(x_0)δ'$. $∴$ We have deduce a contradition.
Since $f$ is continuous, if $f(x) > 0$ then there exists $\varepsilon > 0$ so that $f$ is at a strictly positive distance $d$ from zero in $[x- \varepsilon, x+ \varepsilon] \subset [a,b]$. Thus,
$$ \int_a^bf(t)dt = \int_a^{x-\varepsilon}f(t)dt + \int_{x - \varepsilon}^{x + \varepsilon}f(t)dt + \int_{x + \varepsilon}^bf(t)dt \geq \int_{x - \varepsilon}^{x + \varepsilon}f(t)dt \geq d2\varepsilon > 0. $$
If the first claim is not clear, suppose that $f(x) > 0$ but the claim is false: now for any $n \in \mathbb{N}$, it cannot be that $f(x- \frac{1}{n},x+\frac{1}{n})$ is at positive distance of $0$ and so in particular, some point $x_n \in (x- \frac{1}{n},x+\frac{1}{n})$ must verify $f(x_n) < \frac{1}{n}$. Thus, $x_n \to x$ and so $0 < f(x) = \lim_n f(x_n) \leq \lim_n \frac{1}{n} = 0$ which is absurd.
I think this sums up your argument, without having to handle partitions.