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.

  • $\begingroup$ you are thinking way to hard. It is almost immediate from definition. Recall that convergence a.e = convergence outside a set of measure zero. So outside of a set of measure zero $f_n\rightarrow f$ and outside another set of measure zero $g_n\rightarrow g$.... $\endgroup$ – Yanko Feb 18 '18 at 16:10
  • $\begingroup$ But how can I be sure that $\beta f_n \rightarrow \beta f$ outside a set of $A$, where $\mu(A) = 0$? I only know $f_n \rightarrow f$ outside a set of $A$, where $\mu(A) = 0$ $\endgroup$ – gariban Feb 18 '18 at 16:53
  • $\begingroup$ You should read the definition of convergence a.e. again, do you understand why $f_n\rightarrow f$ implies $\alpha f_n\rightarrow \alpha f$ when the "a.e" is being removed? $\endgroup$ – Yanko Feb 18 '18 at 16:56
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    $\begingroup$ Please add a definition for $f_n\rightarrow f$ a.e, and I will write a full detailed answer for you. $\endgroup$ – Yanko Feb 18 '18 at 18:59
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    $\begingroup$ Thank you. In my book: $f_n$ convergeres to $f$ almost everwhere if $\mu(\{x\in X: lim_{n \rightarrow \infty } f_n = f\}^c ) = 0$. Which is the same you wrote. $\endgroup$ – gariban Feb 18 '18 at 20:20

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.

  • $\begingroup$ Yanko, what if we change inequalities to equalities? I guess there's a problem if $\mu(X)=\infty$? $\endgroup$ – BCLC Apr 8 '18 at 8:24

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