This is false in general.
For instance, if $f$ is any integrable function which is positive everywhere (e.g., $x \mapsto e^{-x^2}$), then this is false. This is because any simple function which is greater than $f$ will have infinite integral.
The requirement for your statement to be true is that $f$ is bounded and nonzero only on a set of finite measure.
This is because we then have that $f \leq M\cdot \mathbf{1}_{f^{-1}(\mathbb{R} \backslash \{0\})}$. Therefore, $M \cdot\mathbf{1}_{f^{-1}(\mathbb{R} \backslash \{0\})}-f \geq 0.$ Letting $g:=M \cdot\mathbf{1}_{f^{-1}(\mathbb{R} \backslash \{0\})}-f$, we can take a sequence $s_n$ of simple functions which converge to $g$ monotonically. Defining $t_n:=M \cdot\mathbf{1}_{f^{-1}(\mathbb{R} \backslash \{0\})}-s_n$, we have that $t_n \to M \cdot\mathbf{1}_{f^{-1}(\mathbb{R} \backslash \{0\})}-g=f$.
By the monotone convergence theorem we have that $\int s_n \to M\mu(f^{-1}(\mathbb{R} \backslash \{0\}))-\int f$, and thus
$$\int t_n=(M\mu(f^{-1}(\mathbb{R} \backslash \{0\}))-\int s_n ) \to (M\mu(f^{-1}(\mathbb{R} \backslash \{0\}))-(M\mu(f^{-1}(\mathbb{R} \backslash \{0\}))+\int f)=\int f.$$
By construction, $t_n:=M \cdot\mathbf{1}_{f^{-1}(\mathbb{R} \backslash \{0\})}-s_n \geq M \cdot\mathbf{1}_{f^{-1}(\mathbb{R} \backslash \{0\})}-g =f,$ and it is now easy to see that the result you state must hold.