$f\in L(\mathbb{R})$, show that if for any open set $G\subset \mathbb{R}$ and $m(G)=1$, we have $\int_{G}f(x)dx=0$, then $f(x)=0$ a.e.$\mathbb{R}$. This is the hint of the problem: $\forall 0<h<1,k\in\mathbb{N}$, let $G:=(x,x+h)\cup(k,k+1-h)$. So, we can get $\int_{k}^{k+1-h}f(t)dt\rightarrow0(k\rightarrow\infty)$. Finally, we obtain $\int_{x}^{x+h}f(t)dt=0$. Why can we get $\int_{k}^{k+1-h}f(t)dt\rightarrow0(k\rightarrow\infty)$.
Generally, if open set $G\subset \mathbb{R}$ has finite Lebesgue measure and  $\int_{G}f(x)dx=0$, can we get the same result?
 A: Suppose $\int_G f(x)dx=0$ for any open set $G$ with $m(G)=t$ where $t$ is a pre-assigned positive number. Let $0 <h <t$.  Then $(x,x+h)\cup (k,k+t-h)$ is an open set whose measure is $h+(t-h)=t$ if $k$ is so large that $(x,x+h)$ and $(k,k+t-h)$ are disjoint.  Also, integrability of $f$ implies that $\int_k^{k+t-h} |f(x)|dx \leq \int_k^{\infty} |f(x)|dx  \to 0$ as $k \to \infty$. Hence, we get $\int_x^{x+h} f(x)dx =0$ whenever $0 <h <t$. Now Lebesgue's Theorem on differentiation of integrals shows that $f=0$ a.e..
A: Another solution is possible, with the weaker hypotesis that $f\in L_{\text{loc}}(\mathbb{R})$:
given $x$, let $A=(x,x+1)$, $B=(y,y+1/2)$ with $y$ such that $A\cap B=\emptyset$.
Now, by hypotesis we have $\int_{(x,x+1/2)}f+\int_{(x+1/2,x+1)}f=\int_Af=0$. Similarly, we get $$\int_{(x,x+1/2)}f+\int_{(y,y+1/2)}f=0=\int_{(x+1/2,x+1)}f+\int_{(y,y+1/2)}f$$
Thus $$\int_{(x,x+1/2)}f=\int_{(x+1/2,x+1)}f=0$$
The same reasoning can be applied iteratively, proving that for every $n\ge 0, 0\le k\le 2^n$,
$$\int_{\left(x+\frac{k}{2^{n}}, x+\frac{k+1}{2^n}\right)}f=0$$
Since $f\in L_{\text{loc}}(\mathbb{R})$, we can apply Lebesgue differentiation theorem and get the result.
