I'm having struggles to prove the following statement because im confused about the notation and im unsure about my approch.
Let $X$ be an arbitrary set and $\mu$ a finite measure on the sigma algebra $\Sigma \subset { 2 }^{ X }$.
Let ${ f }_{ i }\epsilon { L }^{ p }(d\mu )$ with ${ f }_{ i }\rightarrow f\quad in\quad { L }^{ p }(d\mu )$. Define ${ ({ f }_{ i }) }_{ m }:=max(min({ f }_{ i },m),-m)$ show that ${ { f }_{ i } }_{ m }\rightarrow f$ in ${ L }^{ p }(d\mu )$ for ${ i }_{ m }\rightarrow \infty $.
So we want to show that ${ \left\| { { f }_{ i } }_{ m }-f \right\| }_{ p }=0$ if we choose any convergent subsequence that converges to f we get ${ \left\| { { f }_{ i } }_{ m }-f \right\| }_{ p }={ \left\| { { { f }_{ i } }_{ m }-{ f }_{ i } }_{ k }+{ { f }_{ i } }_{ k }-f \right\| }_{ p }\le { \left\| { { { f }_{ i } }_{ m }-{ f }_{ i } }_{ k } \right\| }_{ p }+{ \left\| { { f }_{ i } }_{ k }-f \right\| }_{ p }$ by Minkowski-inequality where the first term goes to zero since it's a Cauchy sequence and the right term also goes to zero since ${ { f }_{ i } }_{ k }\rightarrow f$. So the statement would follow? But what if we have $m<{ f }_{ i }$ than the maximum is m and we can't do the above estimation anymore. Does someone have a tipp on what to do
There are some errors in the statement. You don't start with $(f_i) \in L^{p}$; you start with $f\in L^{p}$ and define $f_i$'s in terms of them. Also limit is taken as $m \to \infty$. You have written $i_m \to \infty$ but there is no $i_m$ here. The statement follows immediately from DCT and the fact that $|f_i| \leq |f|$. [You can use the inequality $(a+b)^{p} \leq 2^{p} (a^{p}+b^{p})$ for $a,b \geq 0$ which gives $|f_i-f|^{p} \leq 2^{p+1} |f|$.].