Let $X$ be a measure space and $E$ be a Banach space. A step function from $X$ to $E$ is a measurable function ($E$ takes the Borel measure) with finite image and whose support has finite measure.
If $f:X\to E $ is an a.e. limit of a sequence of step functions, is it also an a.e. limit of an $L^1$-Cauchy sequence of step functions?
The context for this question is that in Real and Functional Analysis, Lang defines $\mathcal{L}^1$ as the space of a.e. limits of Cauchy sequences of step functions, and I wonder if we can say that it is simply the space of a.e. limits of sequences of step functions.
This is not the case in general. For example, take $X=E=\mathbb R$ (with Lebesgue measure) and let $f_n$ be the characteristic function of the interval $[-n,n]$. Then the constant function $1$ is the pointwise limit of $f_n$, but it clearly cannot be an almost everywhere limit of an $L^1$-Cauchy sequence (since it's not in $L^1$).
Recall that $L^1$ (not $\mathcal L^1$) is a space of equivalence classes of functions rather than a space of actual functions (the equivalence relation being that we identify functions that are equal a.e). However $\mathcal L^1$ is a space of actual functions. Indeed it is the space of functions you get by taking the completion of the step functions with respect to the $L^1$ seminorm before quotienting out by the equivalence relation to get $L^1$. When Lang defines $\mathcal L^1$ as the space of a.e. limits of $L^1$-Cauchy sequences of step functions, he ensures that $\mathcal L^1$ is complete in the $L^1$ (semi)norm, and thus once you quotient you get a space which is a complete normed space.
I hope that clarifies things a bit.