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I am asked to show that the $2-$dimensional Lebesgue measure of the graph of a continuous real function is zero.

Could you give me some hints how I could show it??

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    $\begingroup$ If you know Fubini's theorem, use it: the intersection of the graph with every vertical line has $1$-dimensional measure zero. $\endgroup$ – user147263 Oct 20 '14 at 12:54
  • $\begingroup$ Unfortunately, I don't know Fubini's theorem... $\endgroup$ – Mary Star Oct 20 '14 at 17:47
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Consider the restriction to a compact interval $[a,b]$. Use the uniform continuity of $f$ on the compact interval to show that you can cover that part of the graph by open rectangles of total measure $< \varepsilon$, for any $\varepsilon > 0$.

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    $\begingroup$ No, that corresponds to uniform convergence of a sequence $(f_n)$ of functions to $f$. Uniform continuity is related to continuity (stronger). Recapitulate the definitions. $\endgroup$ – Daniel Fischer Oct 20 '14 at 12:38
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    $\begingroup$ Right. Now if you partition $[a,b]$ into finitely many subintervals of length $< \delta$, can you find a convenient cover of the graph by rectangles whose total measure you can estimate? $\endgroup$ – Daniel Fischer Oct 20 '14 at 12:54
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    $\begingroup$ Consider $u_i = \inf \{ f(x) : x \in [x_i,x_{i+1}]\}$ and $v_i = \sup \{ f(x) : x \in [x_i,x_{i+1}]\}$. $\endgroup$ – Daniel Fischer Oct 20 '14 at 14:00
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    $\begingroup$ Try that. Where does it lead you? (Remember to use the uniform continuity.) $\endgroup$ – Daniel Fischer Oct 20 '14 at 14:14
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    $\begingroup$ That is correct. And if you sum up the volumes of all the rectangles, which bound do you get? $\endgroup$ – Daniel Fischer Oct 20 '14 at 17:53

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