Given: $ f \in L^p([0,1]), p \geq 1, \int_0^1 f(y)\sin{(xy)}dy = 0 \ \forall x \in [0,1] $
Show: $ f=0 $ a.e.
Proof: Since $ |[0,1]| < \infty, f\in L^1 $. By the mean value theorem and DCT, $ g(x) = \int_0^1 f(y)\sin{(xy)}dy $ is differentiable with respect to x, and $ g'(x) = 0 \ \forall x \in (0,1) $. Repeating the process, we see that g has derivatives of all orders, all of which are identically zero. Specifically,
$ \int_0^1 y^{2k}f(y)\sin{(xy)}dy = 0, \ k=0,1,..., \forall x \in (0,1) $.
By the uniform continuity of sin(xy), we can extend this to $ [0,1] $, and $ \int_0^1 y^{2k}f(y)\sin{(y)}dy = 0 $. Since we have that $ f(y)\sin{y} \in L^1 $, Lusin implies given $ \epsilon > 0 \ \exists F \subset [0,1], |[0,1]-F|<\epsilon $, with F compact and $ f(y)\sin{(y)} $ continuous relative to F. By Stone-Weierstrass, there exists a sequence of even polynomials such that $ P_n(x) \rightarrow f(y)\sin{(y)} $ uniformly on F. (We can choose these polynomials to be even by extending $ f(y)\sin{y} $ by even reflection, as in the proof of the Weierstrass Theorem). We obtain from the UCT that
$ 0 = \int_F P_n(x)f(y)\sin{(y)}dy = \lim_{n \rightarrow \infty} \int_F P_n(x)f(y)\sin{(y)}dy = \int_F (f(y)\sin{y})^2dy \Rightarrow f(y)\sin{(y)} = 0 $ a.e. on F $ \Rightarrow f=0 $ a.e. on F.
We now consider the sequence of functions $ f_n = f(y)\chi_{F_{\frac{1}{n}}} $, where $ F_{\frac{1}{n}} $ is the set obtained above by setting $ \epsilon = \frac{1}{n} $. We have that $ f_n \rightarrow f $ in measure, hence a subsequence converges to f uniformly. This implies that $ f=0 $ a.e.
