# showing a function converges pointwise almost everywhere

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:

1. Pointwise convergence of a sequence of functions almost everywhere
2. 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.

• The inequality $\left(1 + \frac{f(x)^{\alpha}}{n^{\alpha}}\right)^{n^{\alpha}} \leq f(x)^{\alpha}$ is false. Take $f(x)=1$, $\alpha>1$. Also, you did not say any thing about the integrability of $f$. Is $f$ itself integrable ? Is it at least bounded ?
– Medo
Mar 16, 2020 at 11:22
• In any case, your pointwise limit should be zero (the identically zero function) wherever $f$ is well-defined. To see this, consider the following exercise: Show that $\lim_{n\rightarrow +\infty}(1+c/n^\alpha)^{1/n}=1$.
– Medo
Mar 16, 2020 at 11:26
• @Medo updated to give details on the function. Thanks I will try your exercise. Can I then say that pointwise convergence implies pointwise almost everywhere convergence? Or do I need to show this separately? Mar 16, 2020 at 11:31
• @Medo any hints on how to prove your exercise? I have tried taking logarithms of both sides. Also the fact it looks somewhat like the exponential limit form. Mar 16, 2020 at 11:52

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 $$0. $$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.
• Thank you that makes sense! I will try to prove the results about $g(t)$ myself for practice. The pointwise argument is clever! Does pointwise convergence imply almost-everywhere convergence then? Mar 16, 2020 at 12:12