# Continuity and non-negative on a closed interval implies integrability?

Suppose $f$ is continuous and non-negative on $[0,1]$ and that $f(0.5)=1$. Then, f is integrable and $\int_0^1 f>0$

This seems very, very intuitive considering the $U(F,P)$ and $L(F,P)$ upper and lower sums respectively are defined $>0$ in this scenario. How can I prove this formally? Is MVT needed with the $f(0.5)$ point?

Approximating $f$ by step functions could help you show that the integral must be positive.
We know from measure theory that if the integral of $f$ on [0,1] is 0, then $f$ must be zero everywhere on [0,1] except for a set of measure zero.
However, the continuity of $f$, along with the fact that $f(0.5) =1$, prevent such a condition from being possible. This would follow from constructing an appropriate sequence of approximating step functions (since their Riemann integrals are easily defined).
See also this question: Show that if $f(x) > 0$ for all $x \in [a,b]$, then $\int_{a}^b f(x) dx > 0$