how to prove the basic property of the Bochner integral in terms of the simple function?

Definition A function $f: E \rightarrow \mathbb{X}$ is called simple if it can be represented as

$$f(\omega)=\sum_{i=1}^k I_{E_i}(\omega) g_i$$

for some finite $k, E_i \in \mathscr{B}$ and $g_i \in \mathbb{X}$.

Definition Any simple function $f(\omega)=\sum_{i=1}^k I_{E_i}(\omega) g_i$ with $\mu\left(E_i\right)<\infty$ for all $i$ is said to be integrable and its Bochner integral is defined as (*):

$$\int_E f d \mu=\sum_{i=1}^k \mu\left(E_i\right) g_i$$

It is not difficult to verify that this definition does not depend on the particular representation of $f$. In particular, the $E_i$ can be chosen without loss to be disjoint. To see this merely observe that if any two sets $E_i$ and $E_j$ in a simple function overlap, which part of the function can be rewritten in an equivalent form using disjoint sets as

$$g_i I_{E_i \cap E_j^c}(\omega)+g_j I_{E_j \cap E_i^c}(\omega)+\left(g_i+g_j\right) I_{E_i \cap E_j}(\omega)$$

with $A^C$ indicating the complement of the set $A$.

The previous definition is extended to a general measurable function from $E$ to $\mathbb{X}$ as follows.

Definition A measurable function $f$ is said to be Bochner integrable if there exists a sequence $\left\{f_n\right\}$ of simple and Bochner integrable functions such that

$$\lim _{n \rightarrow \infty} \int_E\left\|f_n-f\right\| d \mu=0$$

In this case, the Bochner integral off is defined as

$$\int_E f d \mu=\lim _{n \rightarrow \infty} \int_E f_n d \mu$$

To see that this definition has merit first observe from (*) and the triangle inequality that

$$\left\|\int_E f d \mu\right\| \leq \int_E\|f\| d \mu$$

for any simple function $f$.

Question: how to prove that $$\left\|\int_E f d \mu\right\| \leq \int_E\|f\| d \mu$$

My attempt:

1. First, recall the definition of a simple function:

   $f(\omega) = \sum_{i=1}^k I_{E_i}(\omega) g_i$

2. By Definition, the Bochner integral of this simple function is:

   $\int_E f d\mu = \sum_{i=1}^k \mu(E_i) g_i$

3. Now, let's apply the norm to both sides:

   $\left\|\int_E f d\mu\right\| = \left\|\sum_{i=1}^k \mu(E_i) g_i\right\|$

4. By the triangle inequality, we know that the norm of a sum is less than or equal to the sum of the norms:

   $\left\|\sum_{i=1}^k \mu(E_i) g_i\right\| \leq \sum_{i=1}^k \|\mu(E_i) g_i\|$

5. Since $\mu(E_i)$ is a non-negative scalar, we can take it out of the norm:

   $\sum_{i=1}^k \|\mu(E_i) g_i\| = \sum_{i=1}^k \mu(E_i) \|g_i\|$

6. Now, observe that $\|f(\omega)\| = \|\sum_{i=1}^k I_{E_i}(\omega) g_i\| \leq \sum_{i=1}^k I_{E_i}(\omega) \|g_i\|$

I am stuck here.