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I'm stuck in an integration problem. The thing is I don't know how to pick the delta of the statement:

Let $f:[0,2] \rightarrow \mathbb{R}$ continuous, $f(x) \geq0 \quad \forall x\in [0,2]$ y $f(1)=3$. Prove that exists a $\delta>0$ such that $ \int_{0}^{2} f d\alpha \geq 4\delta $

My attempt (or ideas):

For instance, I know implicitly that $f$ is Riemann-Stieltjes Integrable, respect to $\alpha$. Then, that implies that for certain given $\varepsilon>0$ exists $P_\varepsilon \in P[0,2]$ such that $U(f,P_\varepsilon,\alpha)-L(f,P_\varepsilon, \alpha)<\varepsilon$, since $f$ is $\alpha$-integrable. But I can't figure out how to include the hypotheses of f>0 and f(1)=3.

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    What if $\alpha = 0$?2017-02-16
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    It wouldn't be truth, right? Cause $\delta>0$2017-02-16

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