I am studying Real Analysis by Stein and I don't understand how to interpret the word "vanish", which is used informally in the bounded convergence theorem (Theorem 1.4, pg 56). Here is the theorem for reference and the surrounding comments about it.
Suppose that $\{f_n\}$ is a sequence of measurable functions that are all bounded by $M$, are supported on a set $E$ of finite measure, and $f_n(x) \to f(x)$ a.e. $x$ as $n \to \infty$. Then $f$ is measurable, bounded, supported on $E$ for a.e. $x$ and as $n \to \infty$, $$\int |f_n - f| \to 0$$ and $$\int f_n \to \int f$$
Parts of the discussion on the theorem:
- From the assumption one sees at once that $f$ is bounded by $M$ almost everywhere and vanishes outside $E$, except possibly on a set of measure zero.
What does vanish mean? I think that $f$ "vanishes" to 0 outside of $E$. It gets small and "vanishes" to 0, but then I don't understand the exception on why $f$ wouldn't vanish on a set of measure zero so I see my thinking is flawed, but I don't get how. Is my intuition about vanishing completely wrong?
It's because $f_n(x)\to f(x)$ a.e. So lets consider a more specific example. Len $E=[-1,1]\subset \mathbb{R}$ and $\lambda$ the Lebesgue measure on the line. Let's consider $f_n$ are supported on $[-1,1]$ and $f$ is the the pointwise limit and lets say that $f$ vanishes outside of $E$. Then if $g(x):=f(x)$ for $x\in \mathbb{R}\setminus \mathbb{Q\cap E}$ and $g(x) :=f(x)+1=1$ for $x\in \mathbb{Q\cap E}$, then $f_n\to g$ a.e. as well.