Show $ f_n\xrightarrow{L^1}f\iff f_n\to f \text{ in measure} \iff f_n\to f \text{ almost uniformly} \iff f_n\to f \text{ a.e.}$

Question

Consider $(f_n)_n$ an increasing sequence in $\mathcal{L}^1$ and $f\in \mathcal{L}^1$. Show that $$ f_n\xrightarrow{L^1}f\iff f_n\to f \text{ in measure} \iff f_n\to f \text{ almost uniformly} \iff f_n\to f \text{ a.e.}$$

My attempt:

$(1)\Rightarrow (2)$: Let $\varepsilon >0$. From (1) we have$\int |f_n - f| \to 0$ and $\int|f_n-f|\ge \int_{\{ |f_n-f|\ge \varepsilon\}} |f_n-f| \ge \varepsilon \mu\{ |f_n-f|\ge \varepsilon\}$, which proves that $\mu\{ |f_n-f|\ge \varepsilon\} \xrightarrow{n\to\infty} 0. $

$(2)\Rightarrow (3)$: I know that $(f_n)_n$ has a subsequence $f_{n_k}\to f$ almost uniformly. So there is a measurable $E\subseteq \mathbb{R}^d$ and $\varepsilon>0, \varepsilon'>0$ such that $\mu(E^c)\le \varepsilon$ and $|f_{n_k}-f|\le \varepsilon'$ on $E$ for $k\ge N$ for some $N\in\mathbb{N}$. But we have that $n_k \ge k$ for a subsequence and therefore we have, for the same $N\in \mathbb{N}$, that $|f_n-f|\le \varepsilon'$ on $E$ for $n \ge N$. I'm not too sure about the correctness of the proof.

$(3)\Rightarrow (4)$: Pick $E, \varepsilon, \varepsilon'$ as above, then $f_n\to f$ uniformly on $E$, and therefore also pointwise on $E$. Let $D$ be the set of points in $\mathbb{R}^d$ for which $f_n\not\to f$. Then $D\subseteq E^c$ and therefore $\overline{\mu}(D)\le \mu(E^c)\le \varepsilon$, proving that $D$ is a zero set so $f_n \to f$ on $D^c$, or equivalently, $f_n\to f$ a.e.. Is this correct?

$(4)\Rightarrow (1)$: I believe that this follows from monotone convergence: for all $n$ we have that $\int f_n \le \int f < \infty$, as the given sequence is increasing and $f$ is Lebesgue-integrable. All $f_n\in\mathcal{L}^1$, so $f_n\to f$ in $L^1$.

Are my reasonings correct? Are there other (possibly simpler) solution methods?

Thanks.

Answer 1

(1) $\to$ (2): We know $\int_{X} \: |f_{n} - f| < M\epsilon$ for all $n \geq N$. Therefore, by Markov's Inequality, $\{x \in X \: : \: |f_{n} - f| \geq M\} \: \leq \: \frac{1}{M} \cdot \int_{X} |f_{n} - f| < \epsilon$ for $n \geq N$.

(2) $\to$ (3): Let $(f_{n_{j}})_{j = 1}^{\infty}$ be the subsequence that converges almost uniformly. There exists a set $A$ with $m(A) < r$ such that $(f_{n_{j}})_{j = 1}^{\infty}$ converges uniformly on $A^{c}$.

That is, for all $x \in A^{c}$, if $j \geq J$, then $|f_{n_{j}}(x) - f(x)| < \epsilon$. Fix $n > n_{J}$ and we can find $K > J$ such that $n_{K} > n$. Because the sequence is monotone increasing, $$f(x) - \epsilon \: < \: f_{n_{J}}(x) \: \leq \: f_{n}(x) \: \leq \: f_{n_{K}}(x) \: < \: f(x) - \epsilon$$ or equivalently, $|f_{n}(x) - f| < \epsilon$ for $n \geq N$ and all $x \in A^{c}$.

$(3) \to (4):$ For all $n \in \mathbb{N}$, there exists $A_{n}$ with $m(A_{n}) < \tfrac{1}{n}$ such that $(f_{n})$ converges uniformly on $A_{n}^{c}$. In particular, $(f_{n}(x))_{n = 1}^{\infty}$ converges pointwise to $f(x)$ for all $x \in A_{n}^{c}$. If we set $A =\bigcap_{n = 1}^{\infty} \: A_{n}$, then $m(A) = \lim_{n \to \infty} \: m(A_{n}) = 0$ and $f(x)$ converges pointwise for all $x \in A^{c} = \bigcup_{n = 1}^{\infty} \: A_{n}^{c}$.

(4) $\to$ (1): By the Monotone Convergence Theorem, we know $\int_{X} \: f_{n}\to \int_{x} f$. Hence, since $\int_{X} f_{n}, \: \int_{X} \: f < +\infty$, $$\int_{X} \: |f_{n} - f| = \int_{x} (f - f_{n}) = \int_{X} f - \int_{X} \: f_{n} \to 0$$.