Let $f: \mathbb{R} \longrightarrow \mathbb{R}$ s.t $\forall a, \space b \in \mathbb{R}$ with $ a < b $, the function $f$ is Riemann integrable at $[a, b]$ and satisfies: $ \forall \gamma \in \mathbb{R} : \int_{\gamma}^{\gamma +1} f(x) \space dx= 0$
Suppose $f$ is a polynomial function. Prove that $f := 0$.
My intuitive game plan was to consider the interval $[n, n+1]$ for each $n \in \mathbb{N}$. By assumption it implies that there exists $x_{n} \in [n, n+1]$ such that $ f(x_{n}) = 0 $. Now it's only left to show that this property implies $f$ is the zero. I don't see how I can prove it using my Real Analysis 2 tools and would like to get some tips how to proceed from here.
Thanks!
You were almost done. You demonstrated that there is $x_n \in [n, n + 1]$ s.t. $f(x_n) = 0$ for each $n \in \mathbb Z$. This means $f$ has at least countable zeros. So since $f$ is a polynomial, it must be identically zero because non-zero polynomials only ever have finite zeroes.
You can prove this last property by induction on the degree of polynomials $f$. More precisely you can show that any polynomial with (at least) countable zeroes must be identically zero.
Base Case. Suppose $f$ is a degree $0$ polynomial (i.e. it is constant $f(x) = b$) with infinite roots. Then obviously $b = 0$ trivially.
Inductive Case. Now suppose the claim is true for all degrees less than $n$. Consider any degree $n + 1$ polynomial $f$ with countable roots $x_1, x_2, x_3, x_4, \ldots$ Then by the polynomial factorization theorem there is a degree $n$ polynomial $q$ s.t. $f(x) = q(x)(x - x_1)$. Notice all roots (except maybe $x_1$) i.e. $x_2, x_3, x_4, \ldots$ must be roots of $q$. Indeed consider any root $x_i, i \neq 1$ other than $x_1$. Then $q(x_i)(x_i - x_1) = f(x_i) = 0$. And since $(x_i - x_1) \neq 0$ this implies $q(x_i) = 0$. Thus, $q$ is a degree $n$ polynomial with the countable zeroes $x_2, x_3, x_4, \ldots$; so by our inductive hypothesis, $q(x) = 0$ identically. Clearly this means $f(x) = q(x)(x - x_1) = 0$ too, completing the induction.