show that $e^{f(x)}$ is integrable if $f(x)$ is integrable Let $f(x)$ be integrable in the path $[a,b]$, I need to prove that $e^{f(x)}$  is also integrable in $[a,b]$
My attempt is to argue that if $f(x)$ is integrable so $F=\int{f(x)}$ is continous
and than I thought about a way to get relation between $e^F$ to $e^f$
but I guess this is wrong method because I do not see any way to do that.
 A: hint
Let $$g=e^f$$
$f $ is integrable at $ [a,b ]$, then it is bounded $ (|f|\le M )$.
The function $ x\mapsto e^x $ satisfies MVT conditions, so
$$\forall (x,y)\in [a,b]^2$$
$$|g(x)-g(y)|=|e^{f(x)}-e^{f(y)}|$$
$$=|f(x)-f(y)|e^c\le e^M|f(x)-f(y)|$$
from here, you can prove that
$$U(g,P)-L(g,P)\le e^M(U(f,P)-L(f,P))$$
and conclude using Cauchy criterion.
Let $ P=(x_i)_{i=0,n} $ be a partage of $ [a,b]$. then for $ i=0,1,...,n-1 $,
$$I_i=[x_i,x_{i+1}]$$
$$m_i=\inf \{f(x), x\in I_i\}$$
$$M_i=\sup\{f(x),x\in I_i\}$$
and
$$\forall (x,y)\in I_i |f(x)-f(y)|\le M_i-m_i$$
and
$$g(x)\le g(y)+e^M(M_i-m_i)$$
thus
$$\sup \{g(x), x\in I_i\}\le g(y)+e^M(M_i-m_i)$$
and
$$\inf \{g(y),y\in I_i\}\ge \sup\{g(x),x\in I_i\}-e^M(M_i-m_i)$$
A: I shall demonstrate Lebesgue integrability.Since the exponential function defined on the compact interval $[a,b]$ is smooth, it's Lipschitz with Lipschitz constant $K>0$, say.Clearly $e^{f}$ is measurable.Let $s_n$ be a sequence of $L^{1}$ - Cauchy simple functions converging pointwise a.e to $f$. Then $e^{s_n} \to e^f$ pointwise a.e. Also $\int _{[a,b]} |e^{s^m}- e^{s^n}| \leq K \int_{[a,b]} |s_m - s_n|$, implying that $e^{s_n}$ is $L^{1}$ - Cauchy too. Thus $e^f$ is integrable and the integral is the limit of $\int_{[a,b]} e^{s_n}$ as $n \to \infty$.
Note - This pointwise limit of $L^1$- Cauchy simple function approach is often used as a definition of the Bochner integral when the codomain is any Banach space instead of the real numbers.
