Proving an equality of supremums related to the rate function > Given $\sigma > 0$, $$\sup_{\lambda \in \mathbb{R}} \left\lbrace \sigma \lambda\left(\frac{x-m}{\sigma}\right) - \ln \int e^{\lambda \sigma t} \, \mu(dt) \right\rbrace = \sup_{\lambda \in \mathbb{R}} \left\lbrace \lambda\left(\frac{x-m}{\sigma}\right) - \ln \int e^{\lambda t} \,\mu(dt) \right\rbrace$$ I'm trying to prove a property of the rate function $I(x)$ for a probability measure $\mu$; the right hand side of the equation above is $I(\frac{x-m}{\sigma})$ and in order for the result to be true, the equality above must hold. But it's not clear to me how to "pull out" the $\sigma$ from inside of the integral, even with the proviso that only the supremum of the set of numbers need be preserved, since as far as I know there are no actual inequalities, much less equations, properly relating $\int f(x) dx$ to $(\int f(x)^a dx)^{1/a}$ for any real number a that isn't 1.

**Hint:**

$$\sup_{\lambda \in \mathbb{R}} f(\lambda \sigma) = \sup_{\mu \in \mathbb{R}} f(\mu)$$ holds for any function $f$ given that $\sigma>0$.