Prove Integrable Function Property on [0,1]

[mathman]

Let $f(x)$ be an integrable function defined on $[0,1]$ with the following property: $a=inf(f(x))\lt f(x) \lt b=sup(f(x))$. Prove $a\lt \int_0^1f(x)dx \lt b$. It is obviously true, but how does one prove it?

 

[fresh_42]

Let $f(x)$ be an integrable function defined on $[0,1]$ with the following property: $a=inf(f(x))\lt f(x) \lt b=sup(f(x))$. Prove $a\lt \int_0^1f(x)dx \lt b$. It is obviously true, but how does one prove it?

Mean value theorem for integration: https://en.wikipedia.org/wiki/Mean_value_theorem#Mean_value_theorems_for_integration
It requires a continuous function $f(x)$, so one probably has to have a look on the proof. But if $f$ is continuous, we have $\int_0^1f(x)dx = f(\xi)\cdot (1-0)$ with a mean value $a<f(\xi)<b$.

The general version has two functions: $g$ continuous, and $f$ integrable, and says $\int_0^1f(x)g(x)dx=g(\xi)\int_0^1f(x)dx$. Maybe one can find an appropriate $g$ and apply this general version.

 

[member 587159]

Alternatively, write the integral as limit of a sequence of Riemann sums. To get the strict inequality, you will have to play with some $\epsilon$'s though.