Conditional density of Sum of two independent and continuous random variables

Let X and Y be independent and continuous random variables. How do you solve for the conditional density of X+Y given X

The identities $$ P[X+Y\in\mathrm dz\mid X=x]=P[Y\in\mathrm dz-x\mid X=x]=P[Y\in\mathrm dz-x] $$ imply that the conditional density of $X+Y$ given $X$ is $$ f_{X+Y\mid X=x}(z)=f_Y(z-x). $$ Edit: A rigorous approach to find the conditional density $(g_x)_x$ of $X+Y$ conditionally on $X$ is to ask that, for every bounded function $u$, $$ E[u(X+Y)\mid X]=v(X),\qquad v(x)=\int u(z)g_x(z)\mathrm dz. $$ To do that, recall that, by definition, $E[u(X+Y)\mid X]=v(X)$ if and only if, for every bounded function $w$, $E[u(X+Y)w(X)]=E[v(X)w(X)]$, that is, $$ \iint u(x+y)w(x)f_X(x)f_Y(y)\mathrm dx\mathrm dy=\int v(x)w(x)f_X(x)\mathrm dx. $$ By the change of variable $(x,z)=(x,x+y)$, the LHS is also $$ \iint u(z)w(x)f_X(x)f_Y(z-x)\mathrm dx\mathrm dz, $$ hence, by identification, $$ v(x)=\int u(z)f_Y(z-x)\mathrm dz, $$ and, by identification again, everything works fine with $$ g_x(z)=f_Y(z-x). $$ Edit2: A rigorous approach to find the joint density $f$ of $(X+Y,X)$ is to ask that, for every bounded function $u$, $$ E[u(X+Y,X)]=\iint u(z,x)f(z,x)\mathrm dz\mathrm dx. $$ By definition and by the change of variable $(x,z)=(x,x+y)$, the LHS is also $$ \iint u(x+y,x)f_X(x)f_Y(y)\mathrm dx\mathrm dy=\iint u(z,x)f_X(x)f_Y(z-x)\mathrm dx\mathrm dz, $$ hence, by identification, $$ f(z,x)=f_X(x)f_Y(z-x). $$