sequence of step maps which converge almost everywhere

I am trying to learn some measure theory from Lang's Real and Functional Analysis and am having difficulty understanding a claim he makes without proof.

Let $(X, \scr{M}, \mu)$ be a measured space, and let $f:X\to E\;\;$ be a function into a Banach space $E$. Suppose that $\{\varphi_n\}$ is a sequence of step maps from $X$ into $E$ which converges pointwise to $f$ almost everywhere. (Lang calls such functions $\mu$-measurable). Show that then $f$ vanishes outside some $\sigma$-finite subset of $X$.

Any help would greatly be appreciated.

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What is the precise definition of "step function" given in the book? – Jim Belk Jun 8 '11 at 4:25

A step map is defined as follows by Lang: For every step map $\phi$ there is a set $A$ of finite measure and a finite partition $A_1, ..., A_n$ of measurable sets with $$A = \cup_{i=1, ..., n} \; A_i$$ such that $\phi$ is constant on every $A_i$ and zero on the complement of $A$.
Now we have a sequence $(\phi_n)$ of step maps that converges pointwise to a given function $f$ almost everywhere. This sequence has an associated sequence of sets $B_n$ as above, that is every $B_n$ has finite measure and every $\phi_n$ is zero on the complement of $B_n$. By the definition of $\sigma$-finite, the set $$B := \cup_{n \in \mathbb{N}} \; B_n$$ is $\sigma$- finite, since each $B_i$ has finite measure. This means that the pointwise limit $$f = lim_{n \to \infty} \phi_n$$ which is zero on the complement of $B$, is ergo zero outside a $\sigma-$ finite subset of $X$.