Comparing modes of convergence for $(\mathbb{N}, 2^{\mathbb{N}}, \mu ({n})=2^{−n})$

Question

The following is an exercise from Bruckner's Real Analysis:

Let $X =\mathbb{N}$, $M =2^\mathbb{N}$, and $\mu ({n})=2^{−n}$. Determine which of the four modes of convergence coincide in this case.

The four modes are : pointwise convergence, convergence in measure, uniform convergence, convergence in mean (= in $L^1$).

I am starting with an example to reduce number of comparisons : Let $f_n(m)=\dfrac{1}{m^n}$ and $f(x)= \lim_n f_n(x)$. Then $f(1)=1$ and $f(x)=0$ for $x \ne 1$. Then the pointwise convergence does not imply the uniform convergence. The uniform convergence always implies both pointwise convergence and convergence in measure. Sometimes uniform convergence does not imply mean convergence: But I tried the first function and also $f_n =\frac{1}{n} \chi_{[n,2n]}$ and $f(x)= \lim_n f_n(x)$ (in which $\lim_n \int_{\mathbb{N}} f_n d \mu = 0 = f(x)$) but both fail to be counterexamples.

I have no idea for other comparisons especially when convergence in measure comes to play.

Added : the only remaining cases I am struggling are [in measure] implies [uniform] and [in measure] implies [pointwise].

Answer 1

Uniform $\Rightarrow$ all others: You have noted that [uniform] $\Rightarrow$ [pointwise] and [measure] in general.
To see that it implies [mean convergence], we simply note that $$\int|f_n - f| {\mathrm d} \mu \le \|f_n - f\|_{\sup} \int {\mathrm d} \mu = \|f_n - f\|_{\sup},$$ where $\|g\|_{\sup} := \sup_n |g(n)|.$

The above argument works for any finite measure space.

---

[in measure] $\nRightarrow$ [uniform]

Consider $f_n = 2^n\chi_{\{n\}}.$ $f_n$ converges pointwise to $f \equiv 0$.
But $\|f_n - f\|_{\sup} = 2^n \to \infty$ and so, $f_n$ does not converge uniformly.

On the other hand, given any $\epsilon > 0,$ we see that $\mu\{k : |f_n(k) - f(k)| > \epsilon\} \le \mu\{n\} = 2^{-n} \to 0.$
Thus, $f_n$ converges in measure.

---

[in measure] $\Rightarrow$ [pointwise]

Let $(f_n)_n$ be a sequence such that $f_n \to f$ in measure. We show that $f_n \to f$ pointwise.

Fix $m \in \Bbb N$ and let $\epsilon > 0$ be arbitrary. Then, $$\mu\{k : |f_n(k) - f(k)| > \epsilon\} \xrightarrow{n \to \infty} 0.$$ Thus, there exists $N \in \Bbb N$ such that $$\mu\{k : |f_n(k) - f(k)| > \epsilon\} < 2^{-m}$$ for all $n > N$. In particular, $m \notin \{k : |f_n(x) - f(x)| > \epsilon\}$ and so, $|f_n(m) - f(m)| \le \epsilon$ for all $n > N$, as desired.

This argument would work whenever singletons have positive measure.

Answer 2

$\mu$ is a finite measure so uniform convergence implies convergence in mean.

Let $f_n \to f$ in measure.Fix $k$. Suppose $|f_n(k)-f(k)| >\epsilon$. Then $\mu \{i: |f_n(i)-f(i)| >\epsilon\} \geq \mu \{k\}=\frac 1{2^{k}}$. This fails for large enough $n$ so $f_n \to f$ pointwise.

It is easy to give examples where we have pointwise convergence but not uniform convergence.

Any such example also gives an example to show that convergence in measure does not imply uniform convergence.