$f\in L^1(0,1)$ Assume that $\int_{0}^{1}f(t)t^{2n}=0$ for all n=1,2,3... prove that $f=0$ a.e $f\in L^1(0,1)$ Assume that $\int_{0}^{1}f(t)t^{2n}=0$ for all n=1,2,3...  prove that $f=0$ a.e
My work- I can prove this when $f\in C[0,1]$.
$\int f^2= \int|f||f|=\int|f||f-p+p|=\int|f||f-p|+|f||p|$.
I know how to proceed from here.
where P is of the form $p= \sum_{n=0}^{\infty}c_nt^{2n}$.
But my question is what happens when $f\in L^1(0,1)$. Can I still use a approximation theorem?
Here is my work according to @Oliver Diaz.
$A=\{f\in C[0,1]\ such\ that\ f(0)=0\}$ and $B=\operatorname{span}(x^{2n}, n=1,2,3..)$ we can prove $B$ is dense in $A$.
Let us take $f\in A$ then consider, for each $\epsilon>0$ then there exists a $p\in B$ such that $\sup\{f-p\}\leq \epsilon$
$|\int_{0}^{1}f^2|=|\int_{0}^{1}f.f|=|\int_{0}^{1}f.(f-p+p)|\leq|\int_{0}^{1}f.(f-p)|+|\int f.p|=\int_{0}^{1}|f.(f-p)|+|\int f.p|$
here $|\int f.p|=0$ (from the assumption)
then $|\int_{0}^{1}f^2|\leq \int_{0}^{1}|f|.|(f-p)|\leq \int_{0}^{1} M.\epsilon=M.\epsilon $
And then, we can conclude that $f^2=0\ implies\ f=0.$
Now let's take $g\in L^1[0,1]$ , according to @Oliver's fact A is dense in $L^1 norm$.
what I need to show is that if
Here, Im kind of stuck. How do I prove your claim $$\int^1_0 g\mathbb{1}_{(a,b]}=0$$.
 A: This is a sketch of a proof. Try to fill in the missing details.
Let's denote by $m$ the Lebesgue measure on $[0,1]$.
It is enough to assume that, besides being integrable, $f$ is Borel measurable (any Lebesgue measurable function is equal $m$-a.s. to a Borel measurable function).

*

*The space of polynomials in $t^2$ with non zero constant term is dense in $A:=\{f\in C([0,1]): f(0)=0\}$ (uniform norm). This follows from the Stone-Werstrass theorem.


*$A$ is dense in $L_1$, in particular, one can approximate any $\mathbb{1}_{(a,b]}$ by elements of $A$ (in the $L_1$-norm). From this, it follows that
$$\int^1_0 f\mathbb{1}_{(a,b]}=0$$
for all $0\leq a<b\leq 1$.


*The conclusion then should be easy (monotone class arguments, for example).

Edit: sketch of proof of (3): Let $\mathcal{D}$ be the collection of all Borel measurable sets $C$ in $[0,1]$ such that $\int_Cf=0$. As a result of parts (1) and (2),

*

*a. $\mathcal{D}$ contains the collection $\mathcal{C}$ of all  intervals $[a,b]\subset[0,1]$ (notice that if $\int_{(a,b]}f=0$, so is $\int_{[a,b]}f$, for the Lebesgue measure of $\{0\}$ is $0$.)


*b. $\mathcal{D}$ is a $\lambda$-system:
b.1. $[0,1]\in D$ (obvious by (a));
b.2. if $B,C\in \mathcal{D}$ and $B\subset C$, then $C\setminus B\in \mathcal{D}$ (the integrability of $f$ implies $\int_Cf-\int_Bf=\int_{C\setminus B}f$);
b.3.  if $\{B_n:n\in\mathbb{N}\}\subset \mathcal{D}$ and $B_n\subset B_{n+1}$, then $\bigcup_nB_n\in\mathcal{D}$ (dominated convergence)


*c. $\mathcal{C}$ is a $\pi$-system, that is, is closed under intersections.


*d. By Dynkin's monotone class theorem $\sigma(\mathcal{C})\subset \mathcal{D}$. But since the  Borel $\sigma$-algebra, $\mathscr{B}([0,1])$, is generated by $\mathcal{C}$ ( i.e. $\mathcal{B}([0,1])=\sigma(\mathcal{C})$), we have that $\int_Bf=0$ for all $B\in\mathscr{B}([0,1])$.
The punch line: $\{f\neq0\}\in\mathscr{B}([0,1])$  and so, $\int_{\{f\neq0\}}f=0$. This implies that $m(\{f\neq0\})=0$, that is $f=0$ $m$-almost surely.
