Convergence of simple function to measurable function

Question

I am reading the proof of approximation of measurable function by simple functions but I have couple of question. Let me ask these questions:

Theorem 1. Suppose $f:E\to [0,+\infty]$ is a measurable function. Then there exists an increasing sequence of non-negative simple functions $\{\varphi_k\}_{k=1}^{\infty}$ that converges pointwise to $f$, namely, $$\varphi_k(x)\leq \varphi_{k+1}(x) \quad \text{and} \quad \lim\limits_{k\to \infty}\varphi_k(x)=f(x), \ \text{for all} \ x.$$

I understood this theorem completely.

Theorem 2. Suppose $f:E\to [-\infty,+\infty]$ is a measurable function. Then there exists a sequence of simple functions $\{\varphi_k\}_{k=1}^{\infty}$ that satisfies $$|\varphi_k(x)|\leq |\varphi_{k+1}(x)| \quad \text{and} \quad \lim\limits_{k\to > \infty}\varphi_k(x)=f(x), \ \text{for all} \ x.$$

We use the following decomposition of function $f$: $f(x)=f^{+}(x)-f^{-}(x),$ where $$f^{+}(x)=\max\{f(x),0\} \quad \text{and} \quad f^{-}(x)=\max\{-f(x),0\}.$$

Since both $f^+$ and $f^-$ are non-negative, the previous theorem yields two increasing sequences of non-negative simple functions $\{\varphi_k^{(1)}(x)\}_{k=1}^{\infty}$ and $\{\varphi_k^{(2)}(x)\}_{k=1}^{\infty}$ which converge pointwise to $f^+$ and $f^-$, respectively. Then, if we let $$\varphi_k(x)=\varphi_k^{(1)}(x)-\varphi_k^{(2)}(x),$$ we see that $\varphi_k(x)$ converges to $f(x)$ for all $x$. Finally, the sequence $\{|\varphi_k|\}$ is increasing because the definition of $f^+, f^-$and the properties of $\varphi_k^{(1)}$ and $\varphi_k^{(2)}$ imply that $$|\varphi_k(x)|=\varphi_k^{(1)}(x)+\varphi_k^{(2)}(x). \qquad (*)$$

I have only two questions.

1) Why $\varphi_k(x):=\varphi_k^{(1)}(x)-\varphi_k^{(2)}(x)$ converges pointwise to $f=f^+-f^-$? I guess this may not be true. What if $\varphi_k^{(1)}(x)\to +\infty$ and $\varphi_k^{(2)}(x)\to +\infty$ then their difference converges to $\infty-\infty$ which in undefined. I guess that something should be added in order to clarify this.

2) I know that $|f|=f^++f^-$ but I am not able to shat $|\varphi_k(x)|=\varphi_k^{(1)}(x)+\varphi_k^{(2)}(x)$.

I would be very grateful for answers!

Answer 1

1. $\varphi_k^{(1)}(x)$ converges to $f^{+}(x)$ and $\varphi_k^{(2)}(x)$ converges to $f^{-}(x)$. These cannot be $+\infty$ both.
2. For any $x$, either $\varphi_k^{(1)}(x) = 0$ or $\varphi_k^{(2)}(x) = 0$ (or both), because $0 \leqslant \varphi_k^{(1)}(x) \leqslant f^{+}(x)$, $0 \leqslant \varphi_k^{(2)}(x) \leqslant f^{-}(x)$ and at most one of $f^{+}(x)$, $f^{-}(x)$ is nonzero (also of use for 1.).