The notion of a point of vanishing mean oscillation is more general than that of Lebesgue points.

Question

How can I demonstrate that the notion of a vanishing mean oscillation point (VMO) is more general than a Lebesgue point?

The point $y\in\mathbb{R}^n$ is a Lebesgue point if:

$$ \lim_{r\to 0} \oint_{B_r(y)} |f(x)-f(y)|\; dx = 0 $$

The point $y\in\mathbb{R}^n$ is a VMO point (vanishing mean oscillation) if:

$$ \lim_{r\to 0} \oint_{B_r(y)} |f(x)-(f)_{y,r}|\; dx = 0 $$

Here $\oint_U f=\frac{1}{|V|}\int_V f$ is the average integral of $f$ (I don't know the command in LateX to write an integral with a dash in the middle). And $(f)_{y,r} = \oint_{B_r(y)}$.

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I tried to put the limit inside the integral. But I can't see that it generalizes.

Some references

VMO1: p. 2 of https://www.ams.org/journals/tran/1975-207-00/S0002-9947-1975-0377518-3/S0002-9947-1975-0377518-3.pdf

VMO 2: http://en.wikipedia.org/wiki/Bounded_mean_oscillation

Lebesgue point: https://en.wikipedia.org/wiki/Lebesgue_point

Answer 1

Using the Lebesgue differentiation Theorem we know that:

$$ \lim_{r\to 0+}\left(\frac{1}{|B_r(y)|}\int_{B_r(y)} f(x)\, dx \right) = f(y)\; a.e \; y\in \mathbb{R}^n $$

Then

$$ \oint_{B_r(y)} |f(x)-(f)_{r,y} |\; dx \leq \oint_{B_r(y)} |f(x)-f(y)|\, dx + \oint_{B_r(y)} |f(y)-(f)_{r,y} | \to 0$$

Where the first installment of the sum goes to 0 if $y$ is a Lebesgue point and the second, is the Lebesgue Differentiation Theorem above.

Answer 2

$\def\avint{\mathop{\rlap{\raise.15em{\scriptstyle --}}\kern-.13em\int}\nolimits}$ If $y$ is a Lebesgue point, it is easy to see that $f_{y,r} \to f(y)$ when $r \to 0$: \begin{align} |f_{y,r} - f(y)| &= \left| \avint_{B_r(y)} f(x) dx - f(y) \right| \\ &= \left| \avint_{B_r(y)} \left( f(x) - f(y) \right) dx \right| \\ & \leqslant \avint_{B_r(y)} |f(x) - f(y)| dx \\ &\to 0. \end{align}

By triangle inequality, $y$ also satisfies the vanishing mean oscillation condition: \begin{align} \avint_{B_r(y)} |f(x) - f_{y,r}| dx &\leqslant \avint_{B_r(y)} |f(x)-f(y)| dx + \avint_{B_r(y)} |f_{y,r}-f(y)| dx \\ &\leqslant \avint_{B_r(y)} |f(x)-f(y)| dx + |f_{y,r}-f(y)| \\ &\to 0. \end{align}