Suppose that $f \in L^p$ on some measure space and we have $|f| \le C$, where $C$ is a constant. Then how do we show that if $f_n \to f$ in $L^p$, then without loss of generality we can assume $|f_n| \le C$?
I think the trick is to consider $g_n:=-C \vee f_n \wedge C$ instead. But how can I show that $g_n \to f$ in $L^p$ if $f_n$ does?
Taking $g_n$ defined as in your question is a good idea, but let me modify it a bit. Let $\delta > 0$ and $$ g_n = f_n \chi_{\{|f_n| \leq C + \delta\}}.$$
Step 1: For every $\delta > 0$ we have $$ \lim_{n \to \infty} \mu\{|f_n| > C+\delta\} = 0. $$
If not, let $\epsilon > 0$ and a subsequence $f_{n_k}$ such that $\mu\{|f_{n_k}| > C+\delta\} > \epsilon$ for every $k$. Then we have $$ \int |f_{n_k} - f|^p \geq \int_{\{|f_n| > C+\delta\}} |f_{n_k} - f|^p \geq \delta^p \mu\{|f_{n_k}| > C+\delta\} > \delta^p\epsilon, $$ contradicting $f_n \to f$ in $L^p$.
Step 2: Take $g_n = f_n \chi_{\{|f_n| \leq C + \delta \}}$ as above. Then $$ \int |g_n - f_n|^p = \int_{\{ |f_n| > C + \delta \}} |f_n|^p \leq 2^p \int_{\{ |f_n| > C + \delta \}} |f_n-f|^p + \int_{\{ |f_n| > C + \delta \}} |f|^p \\ \leq \lVert f_n - f \rVert_{L^p} + C^p \mu\{|f_{n_k}| > C+\delta\} \to 0.$$
In conclusion, $\lVert g_n - f \rVert_{L^p} \leq \lVert g_n - f_n \rVert_{L^p} + \lVert f_n - f \rVert_{L^p} \to 0$.
This proves that we can assume $|f_n| \leq C + \delta$ for every $\delta > 0$, i.e., $|f_n| \leq C$.