Is post composition continuous on $L^1$? Suppose $\Omega$ is a finite measure space and $f: \mathbb{R}\rightarrow \mathbb{R}$ is a compactly supported continuous function.  Is $$L^1(\Omega, \mathbb{R})\rightarrow L^1(\Omega, \mathbb{R}), g \rightarrow f \circ g$$ a continuous function?  Thank you.
 A: It is continuous.  Let $\varepsilon > 0$.  Since $f$ is compactly supported, and thus uniformly continuous, we can choose
$\delta>0$ such that when $|x-y|<\delta$, $|f(x) - f(y)| < \varepsilon$.  Also let $M = \sup |f(x)|$.  
Now choose $g, h\in L^1(\Omega, \mathbb{R})$.  By the Markov inequality
$$
\left|\{x\in \Omega; |g-h|(x) > \delta\}\right|
\leq \frac{1}{\delta}  \|g-h\|_{L^1}.
$$
Applying the bound $|f\circ g(x) - f\circ h(x)| \leq \varepsilon$ when 
$|g(x) - h(x)| \leq \delta$, and $|f\circ g(x) - f\circ h(x)| \leq 2M$ otherwise, we obtain
$$
\|f\circ g - f\circ h\|_{L^1} = \int_\Omega |f\circ g(x) - f\circ h(x)| \,dx
\leq \varepsilon |\Omega| + \frac{2M}{\delta} \|g-h\|_{L^1}.
$$
Thus if we take $\|g-h\|_{L^1} \leq \varepsilon \delta / M$, then $\|f\circ g - f\circ h\|_{L^1} \leq (|\Omega| + 2) \varepsilon$.  
A: Yes it is. For a contradiction, assume it isn't. Then there is a sequence $f_n \rightarrow g$ in $L^1$ with $f \circ f_n \not \rightarrow f \circ g$.
By switching to a subsequence, we can assume that there is $\epsilon > 0$ with $\Vert f \circ f_n - f \circ g \Vert > \epsilon$ for all $n$ (why?).
Switching to a subsequence again, we can assume $f_n \rightarrow g$ a.e. (why?).
This implies $f \circ f_n \rightarrow f \circ g$ in $L^1$ (why? You will have to use your assumptions on $f$ and on the domain plus some convergence theorem here).
This contradicts the choice of our (sub)sequence.
A: Here is an ugly proof:
Suppose $g_n \to g$ (in $L^1(\Omega)$). Then there is a subsequence such that $g_{n_k}(x) \to g(x)$ for ae. $x$.
Let $\alpha_n = \|f\circ  g - f \circ g_n\|$. We want to show $\alpha_n \to 0$.
Note that there is some $B$ such that $|f(x)| \le B$ for all $x$, hence
$|f\circ  g (x) - f \circ g_n (x)| \le 2 B$ for all $x$.
Choose any subsequence $\alpha_{n_k}$, then there is a sub-subsequence
$g_{n_{k_i}}$ such that $g_{n_{k_i}}(x) \to g(x)$ for ae. $x$. Then the bounded
convergence theorem shows that $\alpha_{n_{k_i}} \to 0$. It follows that $\alpha_n \to 0$.
