$f$ is integrable & continuous over $[a,b]$ , $\int_{a}^{b}f(x)dx \geq 0$ for any subinterval $(\alpha,\beta)$ of $(a,b)$, then $f \geq 0$ in $[a,b]$ Some known things about this problem are: if $f(c) < 0$, $a < c < b$, then $f(x) < f(c)/2$ in some neighborhood of $c$, but I am not exactly sure how to use this to get to my goal of showing that $f \geq 0$ in $[a,b]$. 
 A: By the fundamental theorem of calculus,
$$
          f(x) = \lim_{h\downarrow 0}\frac{1}{h}\int_{x}^{x+h}f(t)dt \ge 0.
$$
A: Hint:
(By contradiction) Suppose that there is $a<c<b$ such that $f(c)<0$, the continuity property of $f$ implies that there exists $\delta>0$ such that $\forall c-\delta < x < c+\delta$, $f(x)<f(c)/2<0$. 
A: Think about what it means for $f(x)$ to be continuous at $x_0$...this means that for each $\varepsilon>0$, $\exists$ some $\delta>0$ such that if $|x-x_0|<\delta$, then $|f(x)-f(x_0)|<\varepsilon$. So, suppose that $f(x_0)<0$ for some $x_0\in (a,b)$. Then continuity implies that there exists some neighborhood $(x_0-\delta,x_0+\delta)$ of $x_0$ such that $f(x)<0$  $\forall x\in (x_0-\delta,x_0+\delta)$. Then $\int_{x_0-\delta}^{x_0+\delta} f(x)dx<0$, a contradiction! Egads! Therefore, $f(x)\geq 0$ on $(a,b)$. The cases for $x_0=a$ or $x_0=b$ are similar, but take a little finagling since we only know the function is continuous on $[a,b]$. Hope this helps!
A: Suppose $f(x) < 0 \in [a,b]$. Then
$$0 >\int_a^bf(x)dx = \lim_{\delta, \eta \to 0} \int_{a+\delta}^{b-\eta}f(x)dx > 0$$ by the continuity of $f$. What gives you a contradiction.
