I'm given a random variable $X$ which takes its values on $[-1,1]$ with $E[X]=0$. I've already proven $e^{\lambda x} \le e^{\lambda^2/2} + x\sinh(\lambda)$ for all $x\in[-1,1], \lambda\ge0$. I'm being asked to show $E[e^{\lambda X}] \le e^{\lambda^2/2}$.
So intuitively the way I prove this is taking the "expectation of both sides." I.e,
$$e^{\lambda X} \le e^{\lambda^2/2} + X\sinh(\lambda) \implies E[e^{\lambda X}] \le E[e^{\lambda^2/2} + X\sinh(\lambda)] = e^{\lambda^2/2} + E[X]\sinh(\lambda) = e^{\lambda^2/2}$$
But I'm not entirely sure if I'm allowed to do this. I know that I'm supposed to be using Lebesgue integrals. How do I use lebesgue integrals to do the equivalent of taking the expectation of both sides? Or is my intuition entirely wrong and I need to use some other approach?
This is totally legitimate. It's just the fact that expectation respects order: if $Y,Z$ are two random variables with $Y \le Z$ almost surely, then $E[Y] \le E[Z]$. Here you have $Y = e^{\lambda X}$ and $Z = e^{\lambda^2/2}+X\sinh(\lambda)$.
The corresponding fact in integral notation is that if $\mu$ is a measure on a set $\Omega$ and $f,g : \Omega \to \mathbb{R}$ satisfy $f(x) \le g(x)$ for almost every $x \in \Omega$, then $\int f\,d\mu \le \int g\,d\mu$. It's one of the first properties you prove when constructing the Lebesgue integral.