What is the interpretation of 'measurable function' here?

Question

On the topic of stochastic integration we have the following.

For Brownian motion $W$ and square-integrable, measurable function $f:[0,1] \rightarrow \mathbb{R}$ one can define a random variable

\begin{align} Z = \int_{0}^{1}f(t)dW_t \end{align}

and this random variable is then such that it follows a normal distribution with mean $0$ and variance $\int_{0}^{1}f(t)^2dt$.

Now my question is on the requirement that the function be measurable. I followed a course treating parts of measure theory. But, I believe I was told that a function can be measurable with respect to a sigma algebra (sigma field). However often I see requirements like the one above, without any mention of a sigma algebra. Can somebody maybe clarify this for me? Is there some convention ?

Answer 1

When "measurable" is used for a function on a real interval without mention of a $\sigma$-algebra, Lebesgue measurable is generally meant.

Answer 2

The usual way to equip $\mathbb R$ and its subsets with a $\sigma$-algebra is to take the Borel $\sigma$-algebra, which is the $\sigma$-algebra generated by all open subsets. This is usually implied if we call a function $f:A\to\mathbb R$, $A\subseteq\mathbb R$ measurable.

There are also other ways to equip it with a $\sigma$-algebra, like the Lebesgue $\sigma$-algebra, or any other $\sigma$-algebra, really. But if it isn't mentioned, then it's the Borel one.