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In some course notes on "evolution equations in probability spaces (etc)" I have the following definition:

A point $z \in \mathbb R^n$ is called the approximate limit of $f$ at $x$ if for every $\epsilon > 0$ the set $\{y \in \Omega : |f(y) - z| \leq \epsilon\}$ has Lebesgue density 1 at $x$. Notation: $\tilde{f}(x) := f(y)$.

I don't understand this definition, what exactly does it say intuitively? The given set also does not depend on $x$... is there something wrong?

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    Yes, there is something wrong: the very first instance of $x$ should be $z$ instead.2010-12-30
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    @Pete L. Clark: Thanks. Could you explain what it means intuitively? What about the notation, what does the $y$ have to do with it, and what $y$ do they mean?2010-12-30
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    Anyway, I would be hard-pressed to answer the question "What exactly does it say intuitively?" in any context. But *roughly*: to say that the limit of $f$ at $x$ is $z$ in the usual sense means that for *all* points $y$ sufficiently close to $x$ the inequality $|f(y) - z| < \epsilon$. The definition is replacing all with "almost all" in the measure theoretic sense.2010-12-30
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    Rather than $x$, $y$ and $z$, I suggest $x_0$, $x$ and $L$. Does that help? Oh, and finally: the bit $\tilde{f}(x) := f(y)$ makes absolutely no sense to me either. But I think it is okay not to understand the notation if you understand the definition!2010-12-30
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    Thanks. I also suppose they mean $\tilde{f}(x) := f(z)$?2010-12-30
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    My best guess at what the notation should be is $\tilde{f}(x) := z$. Thus we redefined the function value at $x$ to be its approximate limit. That makes it *approximately continuous* at $x$.2010-12-30

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I don't know if this will help your intuition, but unwinding the definition a bit, this says that for all $\epsilon\gt0$, for all $\eta\gt0$, there exists $\delta\gt0$ such that in $B_\delta(x)$ you have $|f(y)-z|\leq\epsilon$ outside of a subset of measure less than $\eta\cdot m(B_\delta(x))$. For a fixed $\epsilon$, letting $\eta$ go to zero makes the relative size of the set where $|f(y)-z|\gt\epsilon$ go to zero (by taking sufficiently small neighborhoods of $x$). You get an ordinary limit when you don't have to include the exception of a small subset where $|f(y)-z|$ might be larger than $\epsilon$.

In the simplest case, the limit may exist if you ignore the values of $f$ off a set of measure zero. (E.g., take a continuous function but add $1$ to its values on a countable dense subset of the domain.) In that case you can actually take $\eta=0$. Perhaps more illustrative is the example of the function $f$ on $\mathbb{R}$ that is $1$ in an open interval of length $2^{-n}$ centered at $\frac{1}{n}$ for each positive integer $n$, and $0$ elsewhere. Then the approximate limit of $f$ at $0$ is $0$, but for $\epsilon\lt1$ and $\delta\gt0$, the measure of the subset of $B_\delta(0)$ on which $|f(y)-0|\gt\epsilon$ is positive.

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    Thanks, I was confused by the mistakes in the definition.2010-12-30
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    @Jonas T:You're welcome.2010-12-31