7. Given $f(x)=x^{3}-3 x^{2}+3 x$ has a range of $[a, b]$ on the interval $[a, b](b>a)$. Then the interval $[a, b]$ that satisfies the condition is $\qquad$
7. $[0,1],[0,2],[1,2]$.

From $f^{\prime}(x)=3 x^{2}-6 x+3=3(x-1)^{2}$, we know that except for $x=1$, $f^{\prime}(x)>0$.
Therefore, $f(x)$ is an increasing function on $(-\infty,+\infty)$.
According to the problem, the function takes the minimum value $a$ at $x=a$ and the maximum value $b$ at $x=b$, then
$$
f(a)=a, f(b)=b,
$$

This indicates that $a$ and $b$ are two roots of the equation $f(x)=x$.
Notice that
$$
x^{3}-3 x^{2}+3 x=x \Leftrightarrow x(x-1)(x-2)=0 \text {. }
$$

Solving this, we get $x_{1}=0, x_{2}=1, x_{3}=2$.
Therefore, the required intervals are $[0,1],[0,2],[1,2]$.