About the different types of convergence of Measure Theory

Question

I am studying Measure Theory for the first time and I am having issues to comprehend the different types of convergence and how to study a functions sequence in order to say if it is converges on some type or not.

For example, one of the excercise I have to solve is to determine if $f_n(x):=n^2 e^{-n|x|}$ converges on mean, on measure, almost everywhere and almost uniformly.

My attempt was to prove first that it converges almost uniformly since this is one of the strongest converges (it implies some of the others), but I am not reaching any point, so any advices would be appreaciated :)

Answer 1

The most important first step when dealing with convergence-type exercises is to identify the function the sequence would converge to. In our case, it is not hard to check that that function would be $f \equiv 0.$

It is easy to see that $f_n \to f$ almost surely since the exponential goes to infinity much faster than any power. To establish convergence in mean, we have to compute $$\int_\mathbb{R} f_n d\lambda = 2 \int_0^\infty n^2 e^{-n x} d \lambda = 2 n^2 \frac{e^{-n x}}{-n} \big\vert_0^\infty = 2 n.$$ Of course, this quantity does not converge to $0,$ so the sequence does not converge in mean. We are only left to check the almost uniform convergence. Letting $\varepsilon > 0$ be arbitrary, we see that if $\vert x \vert > \frac{2 \log n - \varepsilon}{n},$ then $\vert f_n(x) \vert < \varepsilon.$ The interval on which this does not hold has measure $2 \frac{2 \log n - \varepsilon}{n},$ which can of course be made as small as one wishes given a sufficiently large $n \in \mathbb{N}.$ It follows that $f_n$ also converges almost uniformly, which also yields the convergence in measure. I hope this helps. :)

Answer 2

• It certainly converges almost everywhere to $0$, and thus in measure, that is for any $x\neq0$, $f_n(x)=n^2e^{-n|x|}\xrightarrow{n\rightarrow\infty}0$ and so $$\lambda(|f_n|>\varepsilon)\xrightarrow{n\rightarrow\infty}0$$ for any $\varepsilon>0$ (here $\lambda$ is Lebesgue's measure.

• As for convergence in say $L_1$ mean, it does not since $$\int^\infty_0n^2e^{-nx}\,dx=-ne^{-nx}\Big|^\infty_0=n\xrightarrow{n\rightarrow\infty}\infty $$

• Uniform convergence occurs in any compact subset $K$ of the real line that does not contain $0$.