I honestly have no idea where to even begin with this question:
Let $f: [0,1]\rightarrow \mathbb{R}$ be bounded and Lebesgue measurable. Suppose that for every $0\leq a<b\leq 1$, we have
$$ \int_a^b f(x)\ \mathsf dx=0. $$
Prove that $f=0$ almost everywhere.
On
Denote $$F(t) = \int_0^t f(x)\,dx \quad 0\le t\le 1.$$
From the condition $F(t)=0$ for any $t\in[0,1]$. Since $F'(t)=f(t)$ a. e., then $f(t) = 0$ a. e.
WLOG $f$ is borel measurable, otherwise replace $f$ with a borel measurable function that equals $f$ almost everywhere. The hypotheses state that $\int f 1_A = 0$ for all intervals $A$. Note that these intervals form a $\pi$-system that generate the borel $\sigma$-algebra. Consider the set of functions for which $\int f g = 0$. It is clearly a vector space, contains the constant function $1$, and if $g_n$ increases to a bounded $g$, then by dominated convergence $\int f g_n \to \int fg$. It follows by the monotone class theorem that $\int f g = 0$ for all bounded borel measurable $g$. Take $g=f$ to conclude that $f=0$ a.e.