I have the following function $n \ln\left(1 + \frac{f(x)^{\alpha}}{n^{\alpha}}\right)$. If $\alpha > 1$ Then I am trying to show that $ \lim_{n \to \infty} \int_{X} f d\mu = 0$. Where by $f$ I mean the given function.
Update:
$f:X \rightarrow [0, \infty]$ is a measurable function such that $ 0 < c:= \int_{X}f d\mu < \infty$
I wish to apply the Dominated Convergence Theorem. However to do this I need to show two things:
- Pointwise convergence of a sequence of functions almost everywhere
- The existence of some other integrable function $g$ such that $|f_{i}| \leq g$ almost everywhere for all $i$.
I'm confused how to show point 1, because i'm not told what the pointwise limit would be, I also need to show that the measure where it does not converge is equal to 0, to meet the almost everywhere requirement.
For point2: I think I have done this by using the fact that $n^{1 – \alpha} \ln \left(1 + \frac{f(x)^{\alpha}}{n^{\alpha}}\right)^{n^{\alpha}} \leq \left(1 + \frac{f(x)^{\alpha}}{n^{\alpha}}\right)^{n^{\alpha}} \leq f(x)^{\alpha}$.
Thanks.
Best Answer
Point-wise convergence follows from the fact that $0 \leq n \ln (1+\frac {f(x)^{\alpha}} {n^{\alpha }}) \leq \frac {f(x)^{\alpha}} {n^{\alpha-1 }} \to 0$.
Now consider the function $g(t)=\frac {\ln (1+t^{\alpha})} t$ defined for $0<t<\infty$. $g$ is continuous, $g(t) \to 0$ as $ t \to 0+$ (since $0 \leq g(t) \leq t^{\alpha -1}$)) and $g(t) \to 0$ as $ t \to \infty$ (as seen by an application of L'Hopital's Rule). These fact imply that $g$ is bounded. If $g(t) \leq M$ for all $t$ then we get $\ln (1+t^{\alpha}) \leq Mt$ from which we get $n \ln (1+\frac {f(x)^{\alpha}} {n^{\alpha }})\leq Mf(x)$. Now we are ready to apply DCT.