Answer a.1: Because $f = f^+ - f^-$ where $f^+ , f^- \geq 0$ and so one only needs the integral on non-negative functions.
Okay, but why exactly does it suffice to deal with $f^+$ and $f^-$ separately? The triangle inequality should be mentioned: for a given $\varepsilon \gt 0$ choose $s^+$ and $s^-$ such that $\|f^\pm-s^\pm\| \lt \varepsilon/2$ and then
$$\|f-(s^+-s^-)\| \leq \|f^+-s^+\|+\|f^--s^-\|\lt \varepsilon$$
by the triangle inequality and the choice of $s^\pm$.
Answer a.2: To prove that one uses that for any non-negative $f$ there exists $s_n$ such that $$\forall \varepsilon > 0 \exists n_0 : n > n_0 \implies |f(x) - s_n(x) | < \frac{\varepsilon}{\mu(\Omega)} \forall x \in \Omega$$ where $s_n \leq s_{n+1}$ and $\displaystyle s_n(x) = \sum_{i=1}^N \alpha_i \chi_{Y_{i,n}}(x)$ and $Y_n := \cup Y_{i,n}$.
Here's a better and more direct way to do it (you don't say how to do it, after all!). Note that the following argument doesn't use $\sigma$-compactness of $\Omega$.
The idea (the basic idea of all of Lebesgue's integration theory!) is to slice the set of values of $f$ into fine strips of height $2^{-n}$:
Assume $f \geq 0$ and put $A_{k,n} = \{x\in \Omega : 2^{-n} k \leq f(x) \lt 2^{-n}(k+1)\}$ and consider $s_n = 2^{-n} \sum\limits_{k=0}^{2^{2n}}k \cdot[A_{k,n}]$. Then $s_n$ approximates $f$ on $\{x \in \Omega\,:\,0 \leq f \lt 2^n + 2^{-n}\}$ up to a precision of $2^{-n}$ pointwise.
Thus $s_n \nearrow f$ pointwise a.e. and since $f \in L^1$ we have $s_n \in L^1$, in particular $s_n$ is supported on a set of finite measure.
By the dominated convergence theorem, then, we have $s_n \to f$ in $L^1$: note that $\|f-s_n\| = \int f-s_n$ and $0 \leq f-s_n \leq f$ while $s_n \to f$ pointwise a.e.
But, as was stated in the exercise you may assume this fact, so there would be no need to spell it out.
Here I think I need to assume $\mu(\Omega) < \infty$ but maybe that follows from what is given in the question.
The space $\mathbb{R}$ with Lebesgue measure is $\sigma$-compact (it is the union of the countably many compact intervals $[-n,n]$, for example) but it is certainly not of finite measure, so no, this does not follow from the assumptions.
Now one has a measurable function with finite (and therefore compact) support.
What? I hope this is a typo. Finite measure certainly doesn't imply compactness, not even boundedness: the set $\bigcup_{n=1}^{\infty} (n, n+2^{-n})$ has Lebesgue measure $1$ is open and unbounded, so...
Next one wants to make that into a continuous function.
b) Recall the Theorem of Lusin and Tietze’s Extension Theorem:
Theorem (Lusin’s Theorem). Let $\Omega$ and $\mu$ as above and $f\colon\Omega\to\mathbb R$ a $\mu$-measurable function with finite support $E$. Then for any $\delta > 0$ there exists a closed set $K\subset E$ such that $μ(E\setminus K) < \delta$ and $f$ is continuous on $K$.
Added: Note that "finite support" should read "support of finite measure" here.
Theorem (Tietze’s Extension Theorem for LCH spaces). If $\Omega$ is a locally compact Hausdorff space and $K \subset \Omega$ compact then any $f ∈ C(K, \mathbb R)$ can be extended to a function on $C_c(\Omega,\mathbb R)$ with supremums norm bounded by $|| f ||_\infty$.
Combine these theorems to show that $C_c(\Omega)$ is dense in $L_1$. You may need that $\mu$ is Radon.
Well, it is certainly a good idea to recall Lusin's theorem and Tietze's extension theorem from time to time, but they are actually not needed here.
We have already shown that each integrable $f \geq 0$ can be approximated in the $L^1$-norm by simple functions with finite support. It remains to show that a characteristic function can be approximated by continuous functions. I'm not spelling the reduction to that fact out, because I should leave something to you.
So let $A \subset \Omega$ be a set of finite measure. Since $\mu$ is a Radon measure on a $\sigma$-compact space (hence it is inner and outer regular on Borel sets of finite measure), we can find a compact subset $K \subset A$ and an open set $U \supset A$ such that $\mu(U \setminus K) \lt \varepsilon$. By Urysohn's lemma we can find a continuous function of compact support $g$ such that $0 \leq g \leq 1$, $g = 1$ on $K$ and $g =0$ outside $U$. This gives that $\int |[A] - g| \leq \mu(U \setminus K) \lt \varepsilon$, hence every characteristic function can be approximated arbitrarily well by continuous functions of compact support.
Answer b:
From a) one has a measurable function $s_n$ with finite support $Y_n$. To make it into a continuous function one can set it to be $0$ on $Y_n \backslash K$ where $K$ is a closed set such that $\mu(Y_n \backslash K) < \delta$ for some $\delta$. Let's call this modified function $\tilde{s_n}$. Then $$ \lim_{n \rightarrow \infty} || \tilde{s_n}(x) - f(x)||_1 = \lim_{n \rightarrow \infty} \int_{Y_n \backslash K} |s_n(x) - f(x)| d\mu = \lim_{n \rightarrow \infty} \int_{Y_n} |s_n(x) - f(x)| d\mu = 0$$
I don't understand this argument at all, I'm afraid. What exactly is $\tilde{s}_n$ and how exactly does that limiting argument work?
Here's a suggestion: use Lusin's theorem to find a closed set of finite measure $K$ such that $\mu(E\setminus K) \lt \delta$ on which $s_n$ is continuous. Use inner regularity of $\mu$ to find a compact set $C \subset K$ with $\mu(C \subset K) \lt \delta$. Apply Tietze's extension theorem to extend the restriction $s_n|_{C}$ to a continuous function of compact support $\tilde{s}_{n}$,close to $s_n$ in the $L^1$-norm.
Your idea is good but needs some finishing touches.
Indeed, for $f* g(x)$ to be non-zero, you have to be able to find some $y\in \mathbb{R}$ such that $f(x-y)\neq 0$ and $g(y)\neq 0$. Then, by definition, $x-y\in \textrm{supp}(f)$ and $y\in \textrm{supp}(g)$, implying that $x\in \textrm{supp}(f)+\textrm{supp}(g)$. Thus, $\textrm{supp}(f*g)\subseteq \overline{\textrm{supp}(f)+\textrm{supp}(g)}$, and we need simply argue that this latter set is bounded.
Indeed, if $y\in \textrm{supp}(f)$ and $z\in \textrm{supp}(g),$ then $||y+z||\leq ||y||+||z||,$ implying that $\textrm{supp}(f)+\textrm{supp}(g)$ is bounded when $\textrm{supp}(f)$ and $\textrm{supp}(g)$ are. Thus, $\textrm{supp}(f*g)$ is a bounded, closed set and hence, compact.
Best Answer
We can ignore the information that $J_\varepsilon$ is a mollifier. All we need is a smooth function with integral one. $J_\varepsilon$ is such a function as proven in a) in the question above.
We will use that $C_c(X)$ is dense in $L^1$ to show that $C_c^\infty(X)$ is also dense in $L^1$ where $X$ is an open subset of $\mathbb{R}$. Let $\epsilon > 0$ and $f \in L^1$. Then by density of $C_c(X)$ there is a $g$ in $C_c(X)$ such that $\| f - g \|_{L^1} < \epsilon$.
Now we need to turn $g$ into a smooth function by convolving it with $J_\varepsilon$. Let $$g_\varepsilon (x) := (J_\varepsilon \ast g ) (x) = \int_\mathbb{R} J_\varepsilon(x - y) g(y) dy$$
Then $g_\varepsilon$ is smooth because $\left ( f \ast g \right )^\prime = f^\prime \ast g = f \ast g^\prime$ and $J_\varepsilon$ is infinitely differentiable.
$g_\varepsilon$ has compact support because if $[-S,S]$ is the support of $g$ and $[-R,R]$ is the support of $J_\varepsilon$ then the support of $J_\varepsilon \ast g$ is contained in $[-S - R, S + R]$ and hence is also compact.
To finish the proof we claim that $\| f - g_\varepsilon \|_{L^1} < \epsilon$:
$$ \| f - g_\varepsilon \| \leq \| f - g \| + \|g - g_\varepsilon \| < \epsilon$$
Where $\| f - g \| < \frac{\epsilon}{2}$ holds because $C_c(X)$ is dense in $L^1$ and $\|g - g_\varepsilon \| < \frac{\epsilon}{2}$ holds because:
$$\begin{align} \|g - g_\varepsilon \|_{L^1} = \int_X \left | g(z) - g_\varepsilon (z)\right | dz &= \int_X \left | g(z) - \int_\mathbb{R} J_\varepsilon(z -y) g(y) dy \right | dz \\ &= \int_X \left | g(z)\int_\mathbb{R}J_\varepsilon(y)dy - \int_\mathbb{R}J_\varepsilon(z -y) g(y) dy \right | dz\\ &\stackrel{(*)}{=} \int_X \left | g(z)\int_\mathbb{R}J_\varepsilon(z - y)dy - \int_\mathbb{R}J_\varepsilon(z -y) g(y) dy \right | dz \\ &= \int_X \left | \int_\mathbb{R} g(z) J_\varepsilon(z - y)dy - \int_\mathbb{R}J_\varepsilon(z -y) g(y) dy \right | dz \\ &\leq \int_X \int_\mathbb{R} | g(z) J_\varepsilon(z - y) |dy - \int_\mathbb{R} | J_\varepsilon(z -y) g(y) | dy dz \\ &= \int_X \int_\mathbb{R} |g(z) - g(y)| J_\varepsilon (z -y) dy dz \end{align}$$
Where the equality marked with (*) holds because the integral is over all of $\mathbb{R}$ so the shift by the constant $z$ doesn't change the integral and $J_\varepsilon$ is even hence $J_\varepsilon (y) = J_\varepsilon (-y)$.
$g$ is continuous and compactly supported hence it is uniformly continuous and so there exists a $\delta$ such that $|g(z) - g(y)| < \frac{\epsilon}{2 \lambda(X)}$ for all $z,y \in X$ hence by choosing $\varepsilon := \delta$ we get
$$ \int_X \int_\mathbb{R} |g(z) - g(y)| J_\delta (z -y) dy dz < \frac{\epsilon}{2} $$
Note that $\epsilon$ and $\varepsilon$ are not the same.