Convergence in $L_p$ and elsewhere

Question

Let $\|f\|_p:=(\int_X|f|^pd\mu)^{1/p}$ and let $L_p$ be the space of (the classes of equivalence of) complex or real measurable functions such that $\int_X|f|^p d\mu<\infty$ exists. In Kolmgorov-Fomin's Элементы теории функций и функционального анализа I find the following interesting properties that are valid for any space $X$ such that $\mu(X)<\infty$:

1. If sequence $\{f_n\}\subset L_2(X,\mu)$ converges with respect to the metric of $L_2(X,\mu)$, it also converges with respect to the metric of $L_1(X,\mu)$ [to the same function, I would say].
2. If sequence $\{f_n\}$ [where I think that it necessary that we intend $\{f_n\}\subset L_2(X,\mu)$] uniformly converges, it also converged with respect to norm $\|\cdot\|_2$ [to the same function, I would say]
3. If sequence $\{f_n\}$ of summable functions [belonging to $L_2(X,\mu)$, I would say, of course] converges with respect to $\|\cdot\|_2$, it also converges in $X$ in measure [to the same function, I would say].
4. If sequence $\{f_n\}$ converges with respect to $\|\|_1$, it is possible to extract a subsequence $\{f_{n_k}\}$ from it that converges almost everywhere [punctually].

From the proofs given by Kolmogorov and Fomin (pp. 387-388 here) for the case of $L_2(X,\mu)$ I am convinced that all that I have written also holds by substituting $L_2$ and $\|\cdot\|_2$ with $L_p$ and $\|\cdot\|_p$, $p\geq 1$. With the precisation that we should have $\{f_n\}\subset L_p(X,\mu)$ at the second point. Is all that I have written correct? Thank you for any answer!!!

Answer 1

Yes, everything is correct. The important things are that if $\mu(X) < \infty$, then we have the inclusions

$$L^q(X,\mu) \subset L^p(X,\mu)$$

for $1 \leqslant p \leqslant q \leqslant \infty$, and by Hölder's inequality

$$\lVert f\rVert_{p}\leqslant \lVert f\rVert_q\cdot \mu(X)^{\large\frac{1}{p}-\frac{1}{q}}$$

these inclusions are continuous. Thus,

1. any convergent sequence in $L^q(X,\mu)$ is also convergent in $L^p(X,\mu)$ for all $1 \leqslant p < q$, and has the same limit (by the continuity of inclusion, but we can also see that by extracting an almost everywhere convergent subsequence).
2. Uniform convergence is slightly stronger than convergence in $L^\infty$ ($L^\infty$-convergence is uniform convergence on the complement of some null set), hence every uniformly convergent sequence converges in all $L^p$ such that $f_n \in L^p(X,\mu)$ for every sufficiently large $n$. For if $N\in\mathbb{N}$ is such that for all $x\in X$ we have $\lvert f_n(x) - f_N(x)\rvert \leqslant 1$ for all $n\geqslant N$, then we have $f_n \in L^p(X,\mu) \iff f_N\in L^p(X,\mu)$ for all $n\geqslant N$, and the sequence $(g_k)$ where $g_k = f_{N+k} - f_N$ is a convergent sequence in $L^\infty(X,\mu)$, hence $g_k\to g$ in all $L^p(X,\mu)$ and $(f_n)_{n\geqslant N}$ therefore converges to $f_N + g$ in all $L^p(X,\mu)$ with $f_N\in L^p(X,\mu)$.
3. The weakest notion of convergence in all the $L^p(X,\mu)$ (where $\mu(X) < \infty$!) is the $L^1$-convergence, so if we see that $L^1$-convergence implies convergence in measure, it follows for all $p\in [1,\infty]$. To see that $L^1$-convergence implies convergence in measure, suppose it were not so, and take a sequence $f_n \xrightarrow{L^1} f$ such that there is a $\delta > 0$ and an $\eta > 0$ with $$\mu\left(\{ x : \lvert f_n(x) - f(x)\rvert > \delta\right) > \eta\tag{$\ast$}$$ for all $n$ [the negation of the definition of convergence in measure gives that inequality for a subsequence $(f_{n_k})$, then we restrict our attention to that subsequence, hence we can assume it holds for all $n$]. Also, using 4., we can assume that $f_n(x) \to f(x)$ pointwise (almost everywhere). Now Egorov's theorem asserts that the convergence is actually uniform on the complement of sets of arbitrarily small measure, i.e. for every $\varepsilon > 0$ there is a measurable $E\subset X$ with $\mu(E) < \varepsilon$ such that the convergence is uniform on $X\setminus E$. Choosing $\varepsilon < \eta$, we obtain a contradiction to $(\ast)$.
4. Since a convergent sequence $(f_n)$ is a Cauchy sequence, we can extract a subsequence such that $\lVert f_m - f_{n_k}\rVert_1 < 2^{-k}$ for all $m \geqslant n_k$. Then $$\sum_{k=0}^\infty \lvert f_{n_{k+1}} - f_{n_k}\rvert$$ is a convergent (in $L^1$) series of non-negative functions, hence it is pointwise convergent everywhere, and attains the value $\infty$ only on a null set $N$. On $X\setminus N$, the series $$\sum_{k=0}^\infty \left(f_{n_{k+1}} - f_{n_k}\right)$$ is absolutely convergent, hence $(f_{n_k}(x))$ converges for all $x\in X\setminus N$.