I want to prove the following result:
If $\left( X, \mathcal{B}, \mu \right)$ is a $\sigma$-finite measure space and $f: X \rightarrow \mathbb{R}$ is a measurable function, then there is a sequence of integrable simple functions which converges pointwise to $f$.
To do so, I am given the hint as contruct $\phi_n: X \rightarrow \mathbb{R}$, as
$$\phi_n = \chi_{X_n} \left( \sum\limits_{k = 1}^{n2^n} \dfrac{k}{2^n} \chi_{f^{-1} \left[ \frac{k}{2^n}, \frac{k + 1}{2^n} \right]} + n \chi_{f^{-1} \left[ n, \infty \right)} \right).$$
Here, $X = \bigcup\limits_{n = 1}^{\infty} X_n$, where $\mu \left( X_n \right) < \infty$ for each $n \in \mathbb{N}$. Now, I can see that each $\phi_n$ is a simple function (since it can take only finitely many values and is defined on measurable sets. Also, it is integrable with the estimate
$$\int\limits_X \phi_n \mathrm{d}\mu = n \mu \left( X_n \right) \left( n 2^{n - 1} + 1 \right).$$
However, what I cannot see (directly) is that why should $\phi_n \to f$ pointwise?
There are some typos in your definition of $\phi_n$. First of all, this works only for non-negative measurable functions $f$ since your $\phi_n$'s are all non-negative. The sum starts from $k=0$. Also, you cannot include both the end points so I will drop the right end-point in this proof.
For any $x$ note that $f(x) <n$ for $n$ sufficienty large. For such $n$ there exists a unique $k$ such that $\frac k {2^{n}} \leq x< \frac {k+1} {2^{n}}$. This implies $k \leq n 2^{n}$. Now $\frac k {2^{n}} \leq x< \frac {k+1} {2^{n}}$ and $\phi_n(x)=\frac k {2^{n}}$. Thes two inequalities give $|f(x)-\phi_n(x)|$ is less than the length of the interval $[\frac k {2^{n}}, \frac {k+1} {2^{n}})$ which is $\frac 1 {2^{n}}$. Hence $\phi_n(x) \to f(x)$. If $f$ is also integrable then so is each $\phi_n$ since $0 \leq \phi_n \leq f$.