Integrability of Composition of continuous and Lebesgue integrable functions Suppose $g:[a,b]\to \mathbb{R}$ is Lebesgue-integrable and $f:\mathbb{R} \to \mathbb{R}$ is continuous, is $f\circ g$ Lebesgue-integrable? 
I am trying to prove this by obtaining simple functions $f_n \to f$ and showing
that $f_n\circ g$ is a cauchy sequence in $\mathcal{L}^1[a,b]$. If we define the 
simple functions as 
$f_n(x) = \begin{cases}
               f(m2^{-n}) & m2^{-n} \leq x < (m+1)2^{-n}, m = 0,1,\cdots, (n-1)2^n\\
               f(-(m+1)2^{-n}) & -(m+1)2^{-n} < x \leq m2^{-n}, m = 0,1,\cdots, (n-1)2^n \\
               n       & |x| \geq n \\
           \end{cases}$
As $g$ is integrable, we have for every $\epsilon > 0$, $\exists M$, 
$\lambda(|g|^{-1}[m,m+1)) < \epsilon$ for all $m > M$.
$\int_a^b |f_n(g(x)) - f_m(g(x))| = \sum_{m > M}\int_{|g|^{-1}[m,m+1)} |f_n(g(x)) - f_m(g(x))| + \int_{|g|^{-1}[0,M)} |f_n(g(x)) - f_m(g(x))|$.
The second integral can be made small as continuity of $f$ and boundedness of the set implies uniform continuity. However, I am unable to make the first term small. Also, I don't know if the claim is correct or if there is a trivial counter-example.
Thanks.

By taking $f(x)=x^2$, your claim would give:
$$ L^1([a,b])\subseteq L^2([a,b]) \tag{1} $$
but such inclusion does not hold - take $[a,b]=[0,1]$ and consider the function that between $\frac{1}{n+1}$ and $\frac{1}{n}$ is defined as $\frac{n}{\log^2(n+1)}$: it is an $L^1$ function, but not an $L^2$ function.
Things change if you further assume that $f$ is Lipschitz continuous or just satisfies some bound of the form $f(x)\ll|x|$, because such conditions give an inequality between the measures of the sets where $f\circ g$ and $g$ are big in absolute value.
