About the proof of Bochner Theorem.

Question

Theorem 8(Bochner) A strongly measurable function $f$ :[0,T]$\rightarrow$ X is summable if and only if t$\rightarrow$ $\left\|f\right\|$ is Lebesgue summable. the proof of "if" part. Let ${f_{n}(t)}$ be a sequence of simple functions strongly convergent to $f(t), \text{a.e}\,t\in [0,T]$. Define $$g_n(t)= \begin{cases} f_n\left( t \right)& \left\| f_n(t) \right\|\leqslant\left\| f(t) \right\|\times(\frac{3}{2})\\ 0& \left\| f_n(t) \right\|>\left\| f(t) \right\|\times(\frac{3}{2})\\ \end{cases}, $$ Then the sequence of simple functions ${g_{n}(t)}$ satisfies $\left\|g_{n}(t)\right\|\leq \left\|f(t)\right\|\times\frac{3}{2}$ and $\lim _{n \rightarrow \infty}\left\|f(t)-g_{n}(t)\right\|=0 \quad\text {a.e.}t\in [0,T].$ Thus, by the lebesgueability of $\left\|f(t)\right\|$, we apply the Lebesgue-Fatou Lemma to the functions $\left\|f(s)-g_{n}(s)\right\| \leqq\|f(s)\|\left(2+2^{-1}\right)$ and obtain $$\int_{0}^{T}\left\|g_{n}(t)-f(t)\right\| d t \rightarrow 0 \text { as } n \rightarrow \infty$$ Why does $g_{n}(t)$ converge strongly to $f(t),\text{a.e.} t\in [0,T]$？