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I have the following question: Let $f=f(x,y)$ be continuous at $(0,0)$ and integrable there. Calculate:$$\lim\limits_{r\to 0^+}\dfrac 1 {\pi r^2}\iint\limits_{|(x,y)|\le r} f(x,y) \, dS$$ Since $f(0,0)$ is not given, I can only conclude from what's given, that $f$ is bounded in the circular vicinity of $(0,0)$, and that doesn't seem to help. Can anyone give me a direction?

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Take $\lambda_r = \sup\limits_{|(x,y)|\leq r} f(x,y)$ and $\mu_r = \inf\limits_{|(x,y)|\leq r} f(x,y)$. Observe we have that for all $r$, $$ \mu_r \leq \frac{1}{\pi r^2}\int_{B(0,r)} f(x,y) \, dx \, dy \leq \lambda_r.$$ Because of continuity we have that $\lambda_r \to f(0,0)$ and $\mu_r \to f(0,0)$ as $r \to 0$. Hence the limit follows.

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  • $\begingroup$ but $f(0,0)$ is not given as i've said. Otherwise it would've been trivial that the integral is $\le f(0,0)*$size of the ball. $\endgroup$ – CodeHoarder Jan 18 '17 at 6:25
  • $\begingroup$ I think you might be confused. The integral is not necessarily bounded above by $f(0,0)\pi r^2$, think about when $f(0,0)$ is a minimum. In regards to not knowing what $f(0,0)$is, it is some value and because of continuity we know that in small neighborhoods around it the function starts approaching that value. $\endgroup$ – Leon Sot Jan 18 '17 at 6:28
  • $\begingroup$ By definition if f is continuous , it's bounded by $f(0,0)+\epsilon$. Therefore, the integral is lower equal to the integral of $f(0,0)+\epsilon$, since that's a constant it equals to $(f(0,0)+\epsilon)\pi r^2$, That's what i've meant. $\endgroup$ – CodeHoarder Jan 18 '17 at 6:31
  • $\begingroup$ @CodeHoarder Yes you can do it that way as well, but remember that is only valid for $r \leq \delta$. Also, in your question you said integrable at $(0,0)$, I assume you meant integrable on some neighbourhood containing $(0,0)$? $\endgroup$ – Leon Sot Jan 18 '17 at 6:34
  • $\begingroup$ Yes, that's what i've meant, thanks. $\endgroup$ – CodeHoarder Jan 18 '17 at 10:30
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HINT If $f(x,y)$ is as nice as you say, it will be approximately constant and equal to $f(0,0).$ in a small enough disk. You can pull $f(0,0)$ out of the integral.

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