# Lebesgue Measures, Limits and Integrals : Prove - If f is in L^1[0,1], then limit the integral over [0,1] of x^n*f = 0 as n goes to infinity.

Prove or disprove the following:

If f is in L^1[0,1], then limit the integral over [0,1] of x^n*f = 0 as n goes to infinity.

I saw a similar example asking to prove that the integral from 0 to 1 of x^2n f(x) dx = 0, and they used algebra of functions generated by {1,x^2}, but we haven't talked about that, so please when you prove or disprove, use basic things we know about Lebesgue measures and integrals, since all integrals here are with respect to Lebesgue measure. If you can't do it this way, then please don't answer my Q. Thanks in advance.

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#### Solution Preview

Proof: (oo represent infinity)

Let g_n=x^n*f. We have the following observations of g_n.

1. |g_n|<=|f|. Because for each x in [0,1], 0<=x^n<=1. So |g_n|=|x^n*f|<=|f|.

2. g_n(x)->0 as n->oo ...

#### Solution Summary

Discusses measurements, limits and integrals.