If $A$ and $B$ are two nonempty subsets of a Banach space $X$, we set $$d(A,B)=\inf\{\|a-b\|:a\in A,b\in B\},$$$$\widehat{d}(A,B)=\sup\{d(a,B):a\in A\}.$$ Thus, $d(A,B)$ is the ordinary distance between $A$ and $B$, and $\widehat{d}(A,B)$ is the non-symmetrized Hausdorff distance from $A$ to $B$.

Let $A$ be a bounded subset of a Banach space $X$. The Hausdorff measure of non-compactness of $A$ is defined by

$\chi(A)=\inf\{\widehat{d}(A,F):F\subset X$ finite$\}$.

Then $\chi(A)=0$ if and only if $A$ is relatively norm compact.

For an operator $T: X\rightarrow Y$, $\chi(T)$ will denote $\chi(TB_{X})$. I have the following question:

Question: Let $T\in \mathcal{L}(X,Y)$ and $S\in \mathcal{L}(Y,Z)$. Then $$\chi(ST)\leq \chi(S)\chi(T).$$

If this question is already known, could you give a proof?

Thank you!