Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $X$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.