I'm being introduced to measure theory, in particular, egoroff's theorem. The version stated in my book considers extended real valued functions and reads as follows:
Let $\Omega \subset \mathbb{R}^n$ be $\mu$ measurable with $\mu(\Omega)<\infty$. Consider a sequence of $\mu$ measurable functions $f_k: \Omega \to \overline{\mathbb{R}}$ converging pointwise $\mu$ almost everywhere to a $\mu$ measurable function $f: \Omega \to \overline{\mathbb{R}}$ with $|f(x)|<\infty$ $\mu$ almost everywhere. Then $$\forall \delta >0 \exists A \subset \Omega \ \mu \ meas.: \mu(A\backslash \Omega) < \delta \ and \ f_k \to f \ unif. \ on \ A$$ Note that if $\mu$ is Radon, then we can choose $A$ to be compact.
The example that came to my mind is probably the first one everyone encounters as a sequence of functions where the convergence is pointwise but not uniform. Take $f_n: x \in [0;1] \mapsto x^n \in \mathbb{R}$ converging pointwise to the zero function on $[0;1)$ and to $1$ for $x=1$. My intuition of the convergence not being uniform, is that the point $x=1$ is the "issue", meaning the convergence is uniform on $[0;1)$. This would be coherent with egoroff's theorem, where $A=\{1\}$ and thus (take $\mathcal{L}^1$ as your measure) $\mathcal{L}^1([0;1]\backslash A)=0< \delta \ \ \forall \delta >0$ and therefore $x^n$ converging uniformly to the zero function on $A=[0;1)$.
My questions are
- Why is my intuition wrong? Thinking about it, $f_k|_{[0;1)}$ still does not converge uniformly to the zero function, or am I mistaken?
- If that is the case, how is egoroff's theorem applicable in this case? What is my A so that the convergence is uniform on A?
Thank you for the help!
Egoroff's theorem says you can take away a set of arbitrarily small positive measure and recover uniform convergence. So you shouldn't necessarily expect to be able to recover it by only deleting the measure-zero set $\{1\}$ from the "problem area".