Let $E, \left \| \right \|$ be a Banach space, $\mathfrak{M}_E$ indicate a family of all nonempty bounded subset of $E$, $\mathfrak{N}_E$ the familly of all relatively compact sets, and $Ker \mu=\{X\in \mathfrak{M}_E$ such that: $\mu(X)=0\}$.

Definition:A mapping $\mu:\mathfrak{M}_E\rightarrow \mathbb R^+$ is said to be a measure of noncompactness in $E$ if it satisfies the following conditions:1) The family $\operatorname{ker} \mu $ is nonempty and $\operatorname{ker} \mu \subset \mathfrak{N}_E$.

2) $X\subseteq Y\Rightarrow \mu (X)\leq \mu (Y)$.

3) $\mu (\bar{X})=\mu (X)$.

4) $\mu (\operatorname{Conv}(X))=\mu (X)$.

5) $\mu (\lambda X+(1-\lambda) Y)\leq \lambda \mu (X)+(1-\lambda) \mu(Y)$ for $\lambda\in [0,1]$.

6)If $\{X_n\}\subset \mathfrak{M}_E^c$, such that $X_{n+1}\subset X_n$ for $n=1,2,...$ and if $\displaystyle\lim_{n\rightarrow \infty} \mu(X_n)=0$ then $X_{\infty}=\bigcap_{n=1}^{\infty}X_n\neq \emptyset$.

Now, I have to prove this this:

Suppose $\mu_1,\mu_2$ are two measures of noncompactness in $E_1,E_2$ respectively. Moreover, assume that the function $F:\mathbb [0,\infty)^2\rightarrow \mathbb R^+$ is a convex function and $F(x_1,x_2)=0\Leftrightarrow (x_1,x_2)=0$. Then $\mu (X)=F(\mu_1(X_1),\mu_2(X_2))$ defines a measure of noncompactness in $E_1\times E_2$ where $X_i$ denote the natural projection of $X$ into $E_i$ for $i=1,2$.

It's okay with 1, 3, 4 and 6 of the definition. *How can we prove 2 and 5?*

EDIT: This result (without proof) comes from : Bana’s, J., Goebel, K., Measures of Noncompactness in Banach Spaces. Lecture Notes in Pure and Applied Mathematics, vol. 60, New York: Dekker 1980