Measures=distributions, what is wrong? We know that in $\mathbb{R}^N$ distributions can be defined as $D(\mathbb{R}^N)=C_c^\infty(\mathbb{R}^N)'$ and Radon finites measures as $m(\mathbb{R}^N)=C_0(\mathbb{R}^N)'$. So, from this definitions we obtain that $D(\mathbb{R}^N) \subset m(\mathbb{R}^N)$.
On the other hand, if $\mu\in m(\mathbb{R}^N)$, the operator $R_\mu (\varphi)=\int_{\mathbb{R}^n}\varphi d\mu$, with $\varphi\in C_c^\infty(\mathbb{R}^N)$, belongs to $D(\mathbb{R}^N)$. Thus, we can identify $m(\mathbb{R}^N) \subset D(\mathbb{R}^N)$.
It is clear that there is something which is wrong in this reasoning, but I don't know where it fails.
Could someone help me please? 
 A: Your first paragraph falsely suggests a containment going in the wrong direction. (Also, using $D$ for distributions is considerably contrary to convention, where some version of "D" would be used for test functions...) Yes, we do have test functions naturally injecting to continuous compact-support functions: $C^\infty_c\to C^o_c$. Taking duals reverses the natural injection:
$$
\hbox{finite Radon measures} \;=\; (C^o_c)^* \;\longrightarrow\; (C^\infty_c)^* \;=\; \hbox{distributions}
$$
Not the other direction.
Note: as @s.harp kindly noted, my previous "inclusion" instead of "natural injection" could too easily give an incorrect impression about whether in $A\subset B$ the topology on $A$ is the subset topology from $B$. In the present case, this is certainly not the case! :)
EDIT: another technical point is that to know that the duals inject uses the density of $C^\infty_c$ in $C^o_o$. When we ask about the density of (finite, Radon...) measures in distributions, we need to specify a topology. If we take the weak dual topology, then in fact (going back to ideas from Garding and others 60+ years ago) test functions are dense in nearly anything with a reasonable topology... although this has some technicalities. So, yes, for specific (not so abstract...) reasons measures are weak-dual dense in distributions, because, in fact, test functions themselves are already weak-dual dense in distributions.
I guess a bit more can be said about these things... :)
