Question

3. Let f : [0, 1] → R be uniformly continuous, so that for every e > 0, there exists >0 such that |x – y < =\f(x) – f(y)] < e for every x, y € [0, 1]. The graph of f is the set Gf = {(x, f(x)) : € [0, 1]}. Show that Gf has measure zero (9 points).

Answer 1

Proof:

Let  $\small \varepsilon >0$ be arbitary and set $\small \eta =\frac{\varepsilon }{2\left ( b-a \right )}$

Since $\small f$ is uniformly continuous on  $\small \left [ 0,1 \right ]$ .

Therefore, there exist $\small \delta > 0$ such that  $\small \left | f\left ( x \right )-f \left ( y \right ) \right |< \eta$ , whenever $\small \left | x-y \right |< \delta$ .

Let  $\small P=\left \{ a=x_{0}< x_{1} < ...< x_{n}=b\right \}$ be any partition whose mesh is $\small < \delta$ .

Define rectangles $\small Q_{0},...,Q_{n-1}$ by $\small Q_{i}=\left [ x_{i} ,x_{i+1}\right ]\times \left [ f\left ( x_{i} \right )-\eta ,f\left ( x_{i} \right )+\eta \right ]$ .

Observe that since $\small \left | x_{i+1}-x_{i} \right |< \delta ,\: \: \: \left | f\left ( x \right )-f\left ( x_{i} \right )\right |< \eta$ , so the portion of the graph of $\small f$ over  $\small \left [ x_{i},x_{i+1} \right ]$ is contained in $\small Q_{i}$ .

Therefore, $\small G_{f}\subset Q_{1}\cup ...\cup Q_{n}$ .

Furthermore,  

$\small \sum_{i=0}^{n-1}\mathrm{vol}\left ( Q_{i} \right )=\sum_{i=0}^{n-1}2\eta \left ( x_{i+1}-x_{i} \right )=2\eta \left ( b-a \right )=\varepsilon$

Since $\small \varepsilon$ is arbitary , this proves that  $\small G_{f}$ has measure zero .