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Recall that the essential range $\operatorname{ess.im} f$ of a measurable function $f \in L^\infty(\mathbb{R})$ is a compact set. Denote by $f_k$ the restriction of $f$ to the interval $[-1/k,1/k]$, and $E_k$ the essential range of $f_k$, and let $E = \bigcap_{k=1}^{\infty} E_k$.

By Cantor's intersection theorem, $E$ is nonempty. This seems dangerously close to assigning function values to $f(0)$, which seems weird to me. Does this thing have a name and where can I find more about it? Or did I screw up somewhere?

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    $\begingroup$ The essential range of a bounded measurable function is the range of the Gelfand transform of the function. So it is natural to think of the essential range as a substitute for the range. $\endgroup$
    – Bill Johnson
    Commented Oct 22 at 19:21
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    $\begingroup$ It should be noted that $E$ may not be a singleton. For example, if $f = 1_{[0, 1]}$, then $E$ is $\{0, 1\}$. $\endgroup$
    – David Gao
    Commented Oct 22 at 19:37
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    $\begingroup$ There is a way of interpreting what $E$ is. Indeed, any character on $L^\infty(\mathbb{R})$ restricts to either the zero functional, or evaluation at some point $x \in \mathbb{R}$, on $C_0(\mathbb{R})$. $E$ is exactly $\{\phi(f)\}$ where $\phi$ ranges over all characters on $L^\infty(\mathbb{R})$ which restrict to evaluation at $x = 0$ on $C_0(\mathbb{R})$. In a sense, this is the “essential” value of $f$ at $0$. I don’t think it has a name or any use though. (Also note that this depends crucially on the topology of $\mathbb{R}$, not just the measure space structure.) $\endgroup$
    – David Gao
    Commented Oct 22 at 19:41
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    $\begingroup$ @LSpice I reckon it is not privileged, any other point admits the same analysis. $\endgroup$
    – Nate River
    Commented Oct 23 at 7:56
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    $\begingroup$ That would not be independent of $x_0$ I think? In the case where $x_0$ is of interest, then OP’s $f_k$ should be the restriction to the intervals $[x_0 - 1/k, x_0 + 1/k]$. $\endgroup$
    – Nate River
    Commented Oct 23 at 8:01

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Geometric measure theory and its many elegant definitions give a way to make sense of these notions. Indeed, many of its tools are well suited for this problem of assigning a pointwise meaning to functions.

Let’s work on an open subset $\Omega$ of $\mathbb R^n$. If the limit

$$L_x := \bigcap_{\delta > 0} \operatorname{ess. im}_{B_\delta (x)} f$$

is unique, then you can indeed assign a value canonically to $f(x).$

This is because $f$ is essentially continuous at $x$, a definition from geometric measure theory. $f$ is said to be essentially continuous at $x$ if there exists a null set $N \subset \Omega$ such that

$$\lim_{y \to x, \, y \in \Omega \setminus N} f(y)$$

exists. Such a limit is necessarily unique if it exists, and course it equals $L_x$ in our case.

One can also look at this from the Lebesgue point perspective. From this point of view, it is also not too surprising that you can canonically assign a value, since if $L_x$ is unique, $f$ is also in the sense of the Lebesgue differentiation theorem, Lebesgue continuous at $x$, in the sense that

$$\lim_{\delta \to 0} \frac{1}{\mu(B_\delta (x))} \int_{B_\delta (x)} |f(y) - L_x| \, dy = 0.$$

Such a limit $L_x$, which exists in fact a.e. for any (locally) $L^1$ function $f$ is known as the sharp representative of $f$ in geometric measure theory, and conditions for its existence at any particular point $x$ are weaker than that which you gave.

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  • $\begingroup$ I'm not sure I know anything about "geometric measure theory" but I needed something stronger than "a.e." and the Lebesgue set for my application. For $x \in \mathbb{R}^d$, I have a closed convex set $Q(x) \subset \mathbb{R}^e$ with nonempty interior. I needed to generalize the notion $f(x) \in Q^{\circ}(x)$. If $f$ is continuous, the meaning is clear. If $f$ is piecewise continuous, I need that $\xi_k \to x \implies \lim f(\xi_k) \in Q^{\circ}(x)$, if the limit exists. The generalization of that notion appears to be $\operatorname{ess val} f(x) \in Q^{\circ}(x)$. $\endgroup$
    – Sébastien Loisel
    Commented Oct 23 at 12:50
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    $\begingroup$ Hm, well i answered your original inquiry to the best of my ability. If you have a more specific question relating to your research goals, you could probably make a new post to ask. @SébastienLoisel $\endgroup$
    – Nate River
    Commented Oct 23 at 18:28

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