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For a locally integrable function $f$ a point $x$ is a Lebesgue point if the integral averages of deviations from $f(x)$ over balls centered at $x$ converge to $0$ as the balls shrink to the point. According to a theorem of Lebesgue almost every point is a Lebesgue point, and obviously every point is Lebesgue if $f$ is continuous. Is the converse true? Are everywhere Lebesgue functions necessarily continuous?

Such questions usually have a "no" answer but I don't see an obvious counterexample. It's easy to change a continuous function into a discontinuous one at a point, that's still Lebesgue at that point, by altering values on a sequence that converges to it. But that would apparently destroy Lebesgueness at the points of the sequence.

Conifold
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    consider $\chi_{[0,1]}:[-1,1] \to \Bbb{R}$ – Loreno Heer Jul 07 '14 at 22:29
  • @sanjab Then $0$ is not a Lebesgue point: $\lim_{h \to 0} \frac{1}{2h} \int_{-h}^h |\chi_{[0,1]}(x)-1| dx = \lim_{h \to 0} \frac{1}{2h} \int_{-h}^0 1 dx = 1/2 \neq 0 $ – Ian Jul 07 '14 at 22:34
  • @Ian ah, I thought the limit will converge to 0, my bad. – Loreno Heer Jul 07 '14 at 22:38
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    There is a weaker definition of Lebesgue point that only requires the value to be equal to the limit of the averages. Then you can get away with jump singularities by assigning the value at them to be the average of the left and right limits, but it won't work with the stronger definition I used. – Conifold Jul 07 '14 at 22:51
  • According to this page an essentially bounded function which is approximately continuous everywhere has every point as a Lebesgue point. – spenceryue May 15 '18 at 12:36

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Let $$f(x) = \sum_{n=1}^\infty (1-4^n|x-2^{-n}|)^+$$ where $a^+=\max(a,0)$. This function consists of triangular spikes of height $1$ at each $x=2^{-n}$, and the width of the spike is $4^{-n}$. It is continuous on $\mathbb R\setminus \{0\}$, but not at $0$. However, $0$ is a Lebesgue point because $$ \int_{-r}^r f(x)\,dx \le \sum_{2^{-n} \le 2r } 4^{-n} = O(r^2) $$ (using the fact that the sum of a geometric series is comparable to its largest term.)