I'm trying to show the following claim:
Let $E$ be a vector bundle on a surface $X$, of slope $\mu(E) = B$ such that it's Harder-Narasimhan factors are $\mu$-semistable of slope $\leq B \in \mathbb{R}$. Then $E$ is $\mu$-semistable.
Here my slope function is $\mu(E) = \frac{c_1(E) . \omega}{\mathrm{rk}(E)}$, and my Harder-Narasimhan filtration is $$ 0 = E_0 \subset E_1 \subset \cdots \subset E_{n-1} \subset E_n = E $$ where the Harder-Narasimhan factors are $F_i = E_i/ E_{i-1}$. The Harder-Narasimhan property gives that the factors are all semistable with slopes $$ \mu(F_1) > \mu(F_2) > \cdots > \mu(F_n). $$ For $E$ to be $\mu$-semistable, I need that for all sub-vector bundles $A \subset E$, the inequality $\mu(A) \leq \mu(E)$ holds (equivalently for all quotients $E \rightarrow Q$, the inequality $\mu(Q) \geq \mu(E)$ holds). By the assumptions in my claim, I know that $\mu(F_i) \leq B$ for all $i$, but I'm not sure how to translate this into a proof of my claim.
My thoughts: I feel like this is really obvious. If I take a sub-vector bundle $A \subset E$, can it somehow be written in terms of the HN factors, and since they all have slope $\leq B$, and $\mu(E) = B$, does this somehow translate into $\mu(A) \leq B$?