The graph of $f:\mathbb R^2\to \mathbb R$ can be embbeded in $\mathbb R^2$ or only in $\mathbb R^3$?

Let $$f:\mathbb R^2\to \mathbb R$$ a continuous function. I'm a bit confuse with something :

The manifold is $$M=\{(x,y,z)\mid z=f(x,y)\},$$ so it's a subset of $$\mathbb R^3$$. This mean that $$M$$ is a submanifold of $$\mathbb R^3$$.

A parametrization of $$M$$ is given by $$\varphi (u,v)=(u,v,f(u,v))$$, and thanks to this, we can define a measure on $$\mathbb R^2$$ (by $$\mu(A\times B)=\int_{A\times B}\sqrt{\det g(u,v)}dudv$$ where $$(g(u,v))_{ij}=\left<\gamma _i,\varphi _j\right>$$ for $$i,j\in \{u,v\}$$). What the meaning of that ? Does it mean that $$M$$ can be embedded in $$\mathbb R^2$$ ?

• $M$ is not a submanifold of $\mathbb{R}^3$. It is a subset of $\mathbb{R}^3$ and it is also a $2$-dimensional manifold over $\mathbb{R}$. This means that locally $M$ is homeomorphic to $\mathbb{R}^2$ (certain open balls in $M$ are homeomorphic to open balls in $\mathbb{R}^2$). It doesn't mean that $M$ can be embedded in $\mathbb{R}^2$. – Yanko Jan 22 at 18:21
• @Yanko : Why it's not a submanifold of $\mathbb R^3$ ? If $g(x,y,z)=f(x,y)-z$, then $M=\{(x,y,z)\in\mathbb R^3\mid g(x,y,z)=0\}$ ? – Dylan Jan 22 at 18:25
• Hmm you might be right about that actually (I think it might depend on $f$ but ignore that). The rest of my comments should be correct. – Yanko Jan 22 at 18:28
• @Yanko: Thank you very much for your comments... very helpful :) – Dylan Jan 22 at 18:29
• @Yanko: Last thing : you agree that to stud $(\mathbb R^2,\mu)$ or $(M,g)$ is really the same right ? For example, the area of a set in $M$ can be given by a the area of a set in $(\mathbb R^2,\mu)$, compute the length of a curve in $(M,g)$ is the same than compute the length of this curve in $(\mathbb R^2,\nu)$ where $d\nu=\sqrt{g(\dot\gamma(t),\dot \gamma (t)) }dt$ ? – Dylan Jan 22 at 18:34

If we have a $$\mathcal{C}^\infty$$ function $$f:\mathbb{R}^2\to\mathbb{R}$$, and we study its graph $$M=\{(x,y,z):z=f(x,y)\}$$ then indeed the space is a smooth manifold with a (global) parametrization $$(x,y)\mapsto (x,y,f(x,y))$$ from $$\mathbb{R}^2\to M$$. $$M\subseteq \mathbb{R}^3$$, and as such it is a submanifold of $$\mathbb{R}^3$$. Not only that: because we have a globally defined coordinate chart $$\mathbb{R}^2\to M$$, we get that $$M$$ is diffeomorphic to $$\mathbb{R}^2$$ and hence embeds in $$\mathbb{R}^2$$. If we replace $$f$$ smooth with $$f$$ continuous, this works except that we only get a homeomorphism, and the resulting surface might not be smooth.
As for the last bit involving the measure, this is saying that because we have a map $$f:\mathbb{R}^2\to M$$ which is a diffeomorphism, we can define a measure on $$A\times B$$ by integrating over the region $$f(A\times B)$$. The meaning of the notation you gave isn't exactly clear to me (what are $$\gamma$$ and $$\phi$$?), so I'll give an explanation in more elementary terms. Recall from multivariable calculus that for $$A,B\subseteq \mathbb{R}$$ $$\text{Area}(f(A\times B))=\int_{f(A\times B)}1\cdot dA=\int_{A\times B}1\cdot \lVert f_x\times f_y\rVert \cdot dA$$ here $$f_x=(\partial_x (x),\partial_x(y),\partial_x(f(x,y))=(1,0,\partial_xf(x,y))$$ and $$f_y=(\partial_y(x),\partial_y(y),\partial_y(f(x,y))=(0,1,\partial_yf(x,y))$$ Then $$f_x\times f_y=\det \begin{bmatrix} i&j&k\\ 1&0&\partial_xf(x,y)\\ 0&1&\partial_yf(x,y) \end{bmatrix}=i(-\partial_xf(x,y))-j(\partial_yf(x,y))+k.$$ $$\lVert f_x\times f_y\rVert=\sqrt{(\partial_x f)^2+(\partial_yf)^2+1}.$$ So, at long last we have defined a measure $$\mu_f(A\times B)=\int_{A\times B} \sqrt{(\partial_x f)^2+(\partial_yf)^2+1}dxdy.$$ To check that this is a measure in the sense of measure theory is not too hard.
$$M$$ is homeomorphic to $$\mathbb{R}^2$$. A canonical homeomorphism is given by $$h : M \to \mathbb{R}^2, h(x,y,z) = (x,y)$$. In particular $$M$$ can be embedded into $$\mathbb{R}^2$$. This does not mean that there exists an embedding which preserves the measure of sets.
In fact $$M$$ is a (topological) submanifold of $$\mathbb{R}^3$$. Define $$\phi : \mathbb{R}^3 \to \mathbb{R}^3, \phi(x,y,z) = (x,y,z-f(x,y))$$. This is a homeomorphism, its inverse being $$\psi(x,y,z) = (x,y,z+f(x,y))$$. We have $$\phi(M) = \mathbb{R}^2 \times \{ 0\}$$ which proves our claim.