$\int_a^b f(x)\mathrm dx=0$ for all $-\inftySuppose $f:\mathbb{R}\to\mathbb{R}$ is Borel measurable. I want to show that $\displaystyle \int_a^b f(x)\mathrm dx=0$ for all $-\infty<a<b<\infty$ with $a$ and $b$ rational implies that $f=0$ a.e. I'm really not sure how to approach this problem; any help would be welcome.
 A: If it's true of all rational $a<b$, then it's true of all real $a<b$.  Pick rational $a_n\uparrow a$ and rational $b_n\downarrow b$ and look at $f\cdot 1_{(a_n,b_n)}$, where $1_S$ is the indicator function of the set $S$.  This converges pointwise to $f\cdot1_{(a,b)}$, and dominated convergence can be used, where the dominating function is $|f\cdot 1_{(a_1,b_1)}|$.
Later edit: Following a suggestion in the comments, I am adding this appendix to my answer.  "Recall" (?) that:
The Lebesgue integral $\int f$ of any non-negative measurable function exists and is either $\infty$ or a finite number.
Any measurable function $f$ can be written as a sum $f^+ - f^-$ of positive and negative parts $f^+$ and $f^-$, where $f^+(x) = f(x)$ if $f(x) \ge 0$ and $f^+(x) = 0$ if $f(x) < 0$, and likewise $f^-(x)= -f(x) > 0$ if $f(x) < 0$ and $f^-(x) = 0$ if $f(x)>0$.  The Lebesgue integral $\int f$ is then defined as $\int f^+ - \int f^-$ provided these are not both infinite.  $f$ is said to be integrable precisely if both of these integrals are finite.  That is the same as saying $\int|f| < \infty$.
