$\int^b_a t^kf(t) dt\,=0$ for all $k \geq 1 \implies f=0$ a.e. 
Let $f\in L^1[a,b]$ satisfying $$\int^b_a t^kf(t) dt\,=0$$  for all positive integer $k$. Show that $f=0$ a.e.

I did a similar problem where $\int^b_a t^kf(t) dt\,=0$ was true for all $k\in \Bbb{N} \cup \{0\}$. This is relatively easy and I did it using Weierstrass approximation theorem, but how to do when $k\in \Bbb{N}$ and k cannot take value $0$.  
 A: The Weierstrass approximation theorem says that the polynomial functions are dense in $C[a, b]$. Since $f \in L^1[a, b]$, it is not necessarily continuous. We should note that $C[a, b]$ is dense in $L^1[a, b]$ before applying the theorem.
To answer your question, there are two approaches. The first is easier. The second is more general.


*

*Consider the function $g(t) = tf(t)$. It is clear that $\int_a^b t^k g(t)\,dt = 0$ for all $k \in \{0, 1, \ldots\}$. By the result you already have, it follows that $g = 0$ [a.e.]. Thus, $f = 0$ [a.e.].

*Use the Müntz–Szász theorem, which generalizes the Weierstrass approximation theorem.

A: Alternative approach using Fourier transforms:
Fix $\xi \in \mathbb{R}$. By the definition of the exponential function, we have
$$\sum_{n=0}^k \frac{(\imath t \xi)^n}{n!} f(t) \to e^{\imath t \xi} f(t)$$
as $k \to \infty$ for any $t \in [a,b]$. Moreover,
$$\left|\sum_{n=0}^k \frac{(\imath t \xi)^n}{n!} f(t) \right| \leq |f(t)| e^{|\xi| \max\{|a|,|b|\}} \in L^1([a,b])$$
for all $k \geq 0$. Therefore, we conclude from the dominated convergence theorem that
$$\hat{f}(\xi := \int_a^b e^{\imath \, t \xi} f(t) \, dt = \lim_{k \to \infty} \sum_{n=0}^k \frac{(\imath \xi)^n}{n!} \int_a^b t^n \cdot f(t) \, dt =\int_a^b f(t) \, dt.$$
Since $\xi \in \mathbb{R}$ is arbitrary, this shows that the Fourier transform $\hat{f}$ of $f \cdot 1_{[a,b]}$ equals the constant $c:= \int_a^b f(t) \, dt$. On the other hand, the Riemann-Lebesgue lemma states that $\hat{f} \in C_{\infty}$; in particular
$$c=\lim_{|\xi| \to \infty} |\hat{f}(\xi)| = 0.$$
Now the uniqueness of the Fourier transform yields $f=0$ (Lebesgue-)almost everywhere on $[a,b]$.
