Suppose $f$ belongs to $L^1[0,1]$ and given any pair of rationals $0\leq p<q\leq 1$ in the case $$\int_p^q fd\mu =0$$ Prove that $f=0$ a.e on [0,1].
I ma preparing for an exam and I came across this question, I need some help. This is what I am thinking.
I have $\int_0^1 |f|d\mu <\infty$ and so $f$ is finite a.e. If I decompose $f$ as $f=f^+-f^-$ where $f^+=max(f(x),0)$ and $f^-=max(-f(x),0)$ and so $f^+$ and $f^-$ are both finite a.e
and let $p=0$, $q=1$.
That means $\int_p^qfd\mu=\int_0^1( f^+-f^-)d\mu=\int_o^1f^+d\mu-\int_0^1f^-d\mu$
but $\int_p^qfd\mu=0$ implies $ \int_0^1f^+d\mu=\int_0^1f^-d\mu$
Hence $f^+=f^-$ a.e and this will mean $f^+=0=f^-$ according to the definition of $f^+$ and $f^-$. To conclude, I will say by monotonicity of the integral, this holds for any other pair of $p,q$.
Does this make sense? if not can someone show me how? Thank you
If $f,g\in L^1(\Omega,\mu)$, then $\int_{\Omega}fd\mu=\int_{\Omega}gd\mu\not\Rightarrow f=g\;a.e.$ A counterexample is given in the comments. However, what is true is that if $\int_{A}fd\mu=\int_{A}gd\mu$ for all $A$ measurable, then $f=g\;a.e.$ (See Folland's Real Analysis proposition $2.23(b)$ for a proof).
With this in mind, I think your idea of a proof can be saved. First, as $f^+$ and $f^-$ are positive functions, the functions $\mu(E)=\int_Ef^+d\mu$ and $\nu(E)=\int_Ef^-d\mu$ are measures on $[0,1]$. Therefore, it suffices to prove that if two measures, $\mu,\nu$ agree on a generating set of a sigma-algebra, then they agree on the entire sigma-algebra (this follows because the set of intervals with rational endpoints generates the Borel sigma-algebra).
A proof of this fact can be found using the Monotone Class Theorem. See, for example, here.