Proof of a known theorem ($f_{xy}=f_{yx}$) It's a pretty famous result but I'm not sure how to prove it.

Let $  f: A \subset \Bbb{R^2} \to \Bbb{R} $ and $(a,b)\in A$ such that $f_{xy}$ and  $f_{yx}$ exist in neighbourhood of $(a,b)$ and are continuous. Show that $f_{xy}(a,b)=f_{yx}(a,b)$.

I don't know much more than the definition $f_{xy} = \lim_{y\to{b}} \frac{f_x(a,y)-f_x(a,b)}{y-b}$.
thanks ahead
 A: We wish to show that $f_{xy}=f_{yx}$ for a continuous function $f(x,y)$ of two variables with continuous second partial derivatives. To do this, we use a helpful lemma. This is known as Clairaut's theorem. 
Lemma: Let $R=[a,b]\times[c,d]$. Then, $$\iint_R f_{xy}\, dA=\iint_R f_{yx}\, dA.$$
Proof: We evaluate each integral. First,
$$\iint_R f_{xy}\, dA=\int_a^b\int_c^d f_{xy}\, dy\, dx =\int_a^b f_x(x,d)-f_x(x,c)\, dx =f(b,d)-f(b,c)-f(a,d)+f(a,c).$$
On the other hand,
$$\iint_R f_{yx}\, dA=\int_c^d\int_a^b f_{yx}\, dx\, dy =\int_c^d f_y(b,y)-f_y(a,y)\, dx =f(b,d)-f(b,c)-f(a,d)+f(a,c).$$
Since these are equal, the lemma holds. We claim the condition from the lemma is equivalent to $f_{xy}=f_{yx}$ for all $(x,y)$. If not, suppose for the sake of contradiction they differ at some point $(x_0, y_0)$, and that 
$$f_{xy}(x_0, y_0)-f_{yx}(x_0, y_0)=\epsilon>0.$$ 
Because both are continuous, there must exist some rectangle $R_0$ with dimensions $\Delta x\times\Delta y$ centered at $(x_0, y_0)$ in which
$$f_{xy}-f_{yx}\geq \frac{\epsilon}{2}.$$
Then,
$$\iint_{R_0} f_{xy}-f_{yx}\, dA \geq \iint_{R_0} \frac{\epsilon}{2}\, dA =\frac{\epsilon}{2}\Delta x \Delta y>0.$$
But from our lemma,
$$\iint_{R_0} f_{xy}-f_{yx}\, dA = \iint_R f_{xy}\, dA-\iint_R f_{yx}\, dA=0,$$
which is a contradiction. Thus, $f_{xy}=f_{yx}$ at every point and they are identical. 
