I'm currently going through all of the theorems in my course on Measure Theory, and when proving the Dominated Convergence theorem I came up with a proof that is very convoluted compared to the way it was proved in class, that doesn't use the Monotone Convergence Theorem or Fatou's Lemma. So I'd be very thankful for some feedback on my proof.
Theorem. Let $\mu$ a Radon measure on $\Bbb{R}^n$ and $\Omega \subseteq \Bbb{R}^n$$\mu$-measurable. Let $g: \Omega \rightarrow [0, \infty]$ be $\mu$- summable ($\int_\Omega|g|\ \text{d}\mu < \infty$) and $f, (f_k)_{k\ge1}: \Omega \rightarrow \overline{\Bbb{R}}$$\mu$-measurable. Suppose $|f_k| \le g$ and $f_k \rightarrow f$$\mu$-a.e as $k \rightarrow \infty$. Then:$$\lim_{k \rightarrow \infty} \int_\Omega |f_k - f|\ \text{d}\mu = 0$$And $$\lim_{k \rightarrow \infty} \int_\Omega f_k\ \text{d}\mu = \int_\Omega f \ \text{d}\mu$$
My Proof Attempt. Let $\varepsilon>0$, $c := \int_\Omega g\ \text{d}\mu < \infty$ and $\forall k\ge 1: A_k := \{x \in \Omega\ |\ \forall n \ge k: |f_k(x) - f(x)| > {\varepsilon \over c}|g(x)|\}$. Since $f_k$ converge pointwise to $f$$\mu$-a.e, we know that $f_k$ converges to $f$ in measure. This means that for any $\varepsilon>0$ and for every $\delta >0$ there exists $K\ge 1$ st. $$\forall k \ge K: \mu(\{x \in \Omega\ |\ |f_k(x)-f(x)| > {\varepsilon \over c}|g(x)|\}) < \delta$$ In other words, $\mu(A_k) < \delta$, which means that $\lim_{k \rightarrow \infty} \mu(A_k) = 0$. Now let $\forall h \geq 1: E_h := \bigcup_{k\ge h} A_k$. $E_h$ is an increasing sequence and therefore we can use the continuity from below of $\mu$ to get: $$\mu(E) := \mu\left(\bigcup_{h\ge1} E_h\right) = \lim_{h \rightarrow \infty} \mu(E_h) \le \lim_{h \rightarrow \infty} \sum_{k\ge h} \mu(A_k) = 0$$The inequality follows from $\sigma$-subadditivity and the last equality follows from $\lim_{k \rightarrow \infty} \mu(A_k) = 0$. Now let $\Omega' := \Omega\setminus E$, then we get:$$\int_\Omega |f_k - f|\ \text{d}\mu = \int_{\Omega'} |f_k - f|\ \text{d}\mu + \int_{E} |f_k - f|\ \text{d}\mu = \int_{\Omega'} |f_k - f|\ \text{d}\mu$$And now by definition of the chosen $\Omega'$, we get my monotonicity of the integral: $$\int_{\Omega'} |f_k - f|\ \text{d}\mu \le \int_{\Omega'} {\varepsilon \over c}|g|\ \text{d}\mu = {\varepsilon \over c} \int_{\Omega}g\ \text{d}\mu = \varepsilon$$Since the $\varepsilon>0$ was arbitray, it follows that $$\lim_{k \rightarrow \infty} \int_\Omega |f_k - f|\ \text{d}\mu = 0$$The second part of the Theorem follows from the triangle inequality: $\forall k\ge1:$$$\left| \int_\Omega f_k - f \ \text{d}\mu\right| \le \int_\Omega |f_k - f|\ \text{d}\mu$$Taking the limit as $k \rightarrow \infty$ we get:$$\lim_{k \rightarrow \infty}\left| \int_\Omega f_k - f \ \text{d}\mu\right| \le \lim_{k \rightarrow \infty}\int_\Omega |f_k - f|\ \text{d}\mu = 0$$Which directly gives us:$$\left|\lim_{k \rightarrow \infty} \int_\Omega f_k - f \ \text{d}\mu\right|= 0 \implies \lim_{k \rightarrow \infty} \int_\Omega f_k - f \ \text{d}\mu = 0$$Since $\forall k \ge 1: |f_k| \le g$, we can take the limit as $k \rightarrow \infty$ on both sides and get $|f| = |\lim_{k \rightarrow \infty} f_k| = \lim_{k \rightarrow \infty} |f_k| \le g$. Therefore we get that $f_k,f$ are both $\mu$-summable (integral bounded by $c<\infty$ by monotonicity) and we can use the linearity of the integral to get:$$0=\lim_{k \rightarrow \infty} \int_\Omega f_k - f \ \text{d}\mu = \lim_{k \rightarrow \infty} \int_\Omega f_k\ \text{d}\mu - \lim_{k \rightarrow \infty}\int_\Omega f\ \text{d}\mu = \lim_{k \rightarrow \infty} \int_\Omega f_k\ \text{d}\mu - \int_\Omega f\ \text{d}\mu \hspace{20pt} \square$$
Remark. My biggest worry for the correctness of this proof is the fact that I don't this I'm using the fact that $g$ bounds the sequence in the first part, and only use the fact that it's summable which implies that it's finite $\mu$-a.e. Any feedback will be greatly appreciated!!