Convergence in $L_p$ and elsewhere 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 $\mu(X)<\infty$:


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*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].

*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]

*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].

*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!!!
 A: 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,


*

*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).

*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 $f_N\in L^p(X,\mu)$.

*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 $(\ast)$.

*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 $x\in X\setminus N$.

