Suppose $f$ is non negative and Lebesgue measurable on $\left[0,1\right]$. If $\int_a^b f(x)dx=b-a$ for all $\left[,\right]\subset \left[0,1\right]$. Show that $f=1$ a.e. on $\left[0,1\right]$.
I tried to break the function in two parts using the indicator function, but someone told me it wasn't the right approach. Instead, he suggested that I should use another function $g=f-1$ and prove it to be zero. My question is, how do we prove it a.e.?
Defining $\nu(A)=\int_A f dx$ you can get a new measure in $[0,1]$ which is "traslation invariant" thanks to the fact that $f \in L^1$.
By unicity of Lebesgue measure in $[0,1]$ we can conclude $f=1$ a.e.
Try this. Suppose $f > 1 + \varepsilon$ on a measurable set $\mathrm{X}$ of positive measure $c > 0.$ There exists an open set $\mathrm{O} = \mathrm{O}_\varepsilon$ containing $\mathrm{X}$ and such that $\lambda(\mathrm{O} - \mathrm{X}) = \lambda(\mathrm{O}) - \lambda(\mathrm{X}) < \dfrac{\varepsilon}{2}c,$ this entails $\lambda(\mathrm{O}) \leq \left( 1 + \dfrac{\varepsilon}{2} \right) c.$ Then $\mathrm{O}$ is the union of a denumerable family of disjoint open intervals $(\mathrm{I}_\alpha).$ Whence, $$\lambda(\mathrm{O}) = \sum_\alpha \lambda(\mathrm{I}_\alpha) = \sum_\alpha \int\limits_{\mathrm{I}_\alpha} f\ d\lambda = \int\limits_\mathrm{O} f\ d\lambda \geq \int\limits_\mathrm{X} f\ d\lambda \geq (1 + \varepsilon) c.$$ Q.E.D.
