4. Find all functions $f: \mathbf{R}_{+} \rightarrow \mathbf{R}_{+}$, such that for all $x, y \in \mathbf{R}_{+}$,
$$
f(x+f(y))=f(x+y)+f(y),
$$

where $\mathbf{R}_{+}$ is the set of positive real numbers.
4. $f(x)=2 x$.

First, we prove: for all $y \in \mathbf{R}_{+}$, we have $f(y) \rightarrow y$.
In fact, from the condition, we get
$$
f(x+f(y))>f(x+y) \text {. }
$$

Therefore, $f(y) \neq y$.
If there exists a positive real number $y$ such that $f(y) < y$,
$$
\begin{array}{l}
f(y) < y,
\end{array}
$$

which is a contradiction.
So, for all $y \in \mathbf{R}_{+}$, we have $f(y)>y$.
For $x \in \mathbf{R}_{+}$, let $g(x)=f(x)-x$. Then
$$
f(x)=g(x)+x \text {, and } g(x)>0 \text {. }
$$

In the original functional equation, let $t=x+y$, we have
$$
f(t+g(y))=f(t)+f(y),
$$

which means
$$
\begin{array}{l}
g(t+g(y))+t+g(y) \\
=g(t)+t+g(y)+y .
\end{array}
$$

Therefore, for any $t>y>0$, we have
$$
g(t+g(y))=g(t)+y .
$$

If there exist $y_{1} 、 y_{2} \in \mathbf{R}_{+}$, such that $g\left(y_{1}\right)=g\left(y_{2}\right)$,
then, for any $t>\max \left\{y_{1}, y_{2}\right\}$, we have
$$
\begin{array}{l}
g(t)+y_{1}=g\left(t+g\left(y_{1}\right)\right) \\
=g\left(t+g\left(y_{2}\right)\right)=g(t)+y_{2} .
\end{array}
$$

Therefore, $y_{1}=y_{2}$. Thus, $g$ is injective.
Let $u 、 v$ be any positive real numbers, $t>u+v$. Then
$$
\begin{array}{l}
g(t+g(u)+g(v))=g(t+g(u))+v \\
=g(t)+u+v=g(t+g(u+v)) . \\
\text { Thus, } t+g(u)+g(v)=t+g(u+v) . \\
\text { Therefore, } g(u)+g(v)=g(u+v) .
\end{array}
$$

Thus, $t+g(u)+g(v)=t+g(u+v)$.
Therefore, $g(u)+g(v)=g(u+v)$.
Since $g(v)$ is a positive real number, therefore, $g$ is strictly increasing.
Since $g(t)+y=g(t+g(y))$
$$
=g(t)+g(g(y)) \text {, }
$$

we have $g(g(y))=y$.
If there exists an $x \in \mathbf{R}_{+}$, such that $g(x) \neq x$. By the monotonicity of $g$: when $x>g(x)$, $g(x)>$ $g(g(x))=x$, which is a contradiction; when $x<g(x)$, $g(x)<$ $g(g(x))=x$, which is a contradiction.

Thus, $g(x)=x, f(x)=g(x)+x=2 x$ $\left(x \in \mathbf{R}_{+}\right)$.