5-4

Question 5.4 
If f\inL(0,1),show that x^k f(x)\in L(0,1) for k =1,2,3,,,,,,and that\int_{0}^{1} x^k f(x) dx\rightarrow0. 
Solution: 
Given that f\inL(0,1).Note that x^k
 is continuous and measurable on (0,1).So x^k f(x)
 is measurable on(0,1), for all k\in\Nu. 
For x\in(0,1), x^k\leq1,so x^k f(x)
\leq f(x) for all k\in\Nu. 
Hence \int_{0}^{1} x^k f(x) dx\leq \int_{0}^{1} f(x) dx<\infty.Therefore x^k f(x)\in L(0,1)for all k\in\Nu. 
Because \left\vert f(x) \right\vert<\infty a.e. since f\inL(0,1),so x^k f\rightarrow0 a.e on (0,1). 
Then since \left\vert x^k f(x) \right\vert<f(x)
,and \left\vert f \right\vert\in L(0,1), By Lebesgue’s Dominated Convergence Theorem,\int_{0}^{1} x^k f(x) dx\rightarrow\int_{0}^{1} 0 dx=0. 
(proved)