Let $(X, \mathcal{E}, \mu ) $ be a measurespace. Let $(f_n)_{n\in\mathbb{N}}$, $f$, $(g_n)_{n\in\mathbb{N}}$ and $g$ $\in \mathcal{M}(\mathcal{E})$. Let $\alpha, \beta \in \mathbb{R}$. Show, if:
$f_n \rightarrow f$ $\mu$-a.e and $g_n \rightarrow g$ $\mu$-a.e $\hspace{1mm}$ then: $\hspace{1mm}$ $\alpha f_n + \beta g_n \rightarrow \alpha f + \beta g$ $\mu$-a.e..
I dont want a solution for this exercise, but just some help or a hint. My idea is showing this $\sum_{n=1}^\infty \int_X |(\alpha f_n + \beta g_n) - (\alpha f + \beta g)| <\infty $. But I'm stuck, and I don't know how to prove this.
Any help is appreciated.
By assumption and your definition we have
$A=\{x\in X: \lim_{n\rightarrow\infty} f_n(x)\not=f(x)\}$
$B=\{x\in X: \lim_{n\rightarrow\infty} g_n(x)\not=g(x)\}$
Are sets of measure zero. For all $x\not\in A$ we have $\lim_{n\rightarrow\infty} f_n(x) = f(x)$ and so $\lim_{n\rightarrow\infty} \alpha f_n(x)=\alpha f(x)$. The same with $g$ and $x\not\in B$.
Now let $C=A\cup B$ we have for all $x\not\in A$ that both $\lim_{n\rightarrow\infty} \alpha f_n(x)=\alpha f(x)$ and $\lim_{n\rightarrow\infty} \alpha g_n(x)=\alpha g(x)$ which implies that $\lim_{n\rightarrow\infty} \alpha f_n+\beta g_n(x)=\alpha f(x)+\beta g(x)$.
Now $\mu(C)\leq \mu(A)+\mu(B)=0+0=0$. Hence by definition $\{x\in X:\lim_{n\rightarrow\infty} \alpha f_n+\beta g_n(x)\not =\alpha f(x)+\beta g(x)\}$ is a subset of $C$ which is of measure zero and so it is also of measure zero.