Let $f:[a,b] \rightarrow\mathbb{R}$ be Riemann integrable and U=$\{x \in [a,b]|f(x)\neq0\}$ have zero measure. Prove that $\int_a^bf(x)=0.$
I'm a bit lost since $U$ isn't closed and thus isn't compact, so how do I build a partition for $[a,b]$?
Here's my attempt at it:Let $U = \bigcup_{n\geq1}E_n$, where $E_n=\{x\in[a,b]| |f(x)| \geq 1/n\}$. Since $E_n \subset U$, $m(E_n)=0, \forall n$. Note also that $E_n$ is closed and thus compact.
Thus given $\epsilon >0$, choose $n$ such that $\frac{b-a}{n} < \epsilon$. Since $m(E_n) = 0$ there exists open interval $I_1,I_2,...$ such that
$E_n \subset \bigcup I_j, \Sigma|I_j|<\frac{\epsilon}{2(M-m)}$, where $M=\sup f,m=\inf f$.
Since $E_n$ is compact there exists a finite subcover $I_1,...,I_n$. Take the partition $P$ such that it's points are extremes of $I_1\cap[a,b],...,I_n\cap[a,b]$.
4Therefore:
$$S(|f|,P)=\Sigma M_i(x_i-x_{i-1})+\Sigma M_i (x_i-x_{i-1})$$
Where the first sum is taken on intervals that do not intercept $E_n$ and the second on intervals that do. Thus
$$S(|f|,P)\leq\Sigma(1/n)(x_i-x_{i-1})+\Sigma(M-m)(x_i-x_{i-1})$$
$$S(|f|,P)\leq\frac{b-a}{n}+(M-m)\frac{\epsilon}{M-m}$$
$$S(|f|,P)<\epsilon/2+\epsilon/2=\epsilon$$
Thus $\overline\int|f|=0\implies\overline\int f=0\implies\int f =0$
Is this ok?
Sketch: I assume $f$ is Riemann integrable on $[a,b]$ and you want to stay within this context. Let $P$ be a partition of $[a,b].$ Then $P$ induces subintervals $I_k, k =1,2,\dots ,n.$ Because $U$ has zero measure, $U$ does not contain any $I_k.$ So for each $k$ there exists $c_k\in I_k$ such that $f(c_k)=0.$ It follows that
$$\sum_{k=1}^{n} f(c_k)\Delta x_k=0.$$
Since $\int_a^b f$ is the limit of such sums (as the mesh size of $P \to 0),$ we have $\int_a^b f=0.$