Let $(S, \mathcal A, \mu)$ be a measure space and $x : S \longrightarrow X$ be a function to a Banach space $X.$ Then $x$ is Bochner integrable if there exists a sequence of simple measurable functions $(x_n)$ such that $\lim\limits_{n \to \infty} \|x_n(s) - x(s)\| = 0$ a.e. and $$\lim\limits_{n \to \infty} \int_{S} \|x_n - x\|\ d\mu = 0.$$
Now in my lecture note the following facts have been left as exercises $:$
$(1)$ If $(x_n)$ is a sequence of simple measurable functions satisfying the above conditions then $\lim\limits_{n \to \infty} \int_{S} x_n$ exists.
$(2)$ If $(x_n)$ and $(y_n)$ be two sequences of simple measurable functions satisfying the above conditions then $$\lim\limits_{n \to \infty} \int_{S} x_n\ d\mu = \lim\limits_{n \to \infty} \int_{S} y_n\ d\mu.$$
My Attempt $:$ For $(1),$ I am trying to show that the sequence $\left ( \int_{S} x_n\ d\mu \right )$ is Cauchy. Here's how I have done it.
$$\left \|\int x_n - \int x_m \right \| \leq \int \|x_n - x\| + \int \|x_m - x\|\ \to 0$$ and now since $X$ is a Banach space we are through. The second one can be tackled in a similar fashion.
But my question is that what's the necessity of the first condition in the definition i.e. why are we assuming that $x$ is strongly measurable?
Any suggestion in this regard would be much appreciated. Thanks for investing your valuable time in reading my question.