step function vs integrable function The Problem:
Let f be an integrable function on $\mathbb{R}$ with Lebesgue measure. Thern for any $\epsilon >0$ there is a finite collection of intervals $E_1, . . . , E_k$ (that is, $E_i = [a_i, b_i])$ and constants $a_1, . . . , α_k$, such that $ \int_{\mathbb{R}}{\left| f(x)-\sum_{i=0}^{k}{\alpha_i 1_{E_i}(x)} \right|} dx <\epsilon$
My problem is the start. How should I begin proving it? How can I say that these intervals and constant exist?
Thank you for any help
 A: Hints:


*

*Start by reducing the problem to positive functions (that is, show that if it holds for every positive function, then it also holds for any integrable function).

*Reduce the problem to positive functions with compact support.

*Reduce the problem to simple functions, and in particular to an indicator function.

Added (it should be safe to say a few more words now) for ease of notation, call a simple function $h=\sum \alpha_j1_{E_j}$ where each $E_j$ is an interval an R-simple function.


*

*Decompose $f=f_+-f_-$. If $h_+,h_-$ are R-simple such that
$$\int_\mathbb{R}|f_+-h_+|dx<\frac{\epsilon}{2};\quad \int_\mathbb{R}|f_--h_-|dx<\frac{\epsilon}{2}$$
then
$$\int_\mathbb{R}|f-(h_+-h_-)|dx \leq \int_\mathbb{R}|f_+-h_+|dx + \int_\mathbb{R}|f_--h_-|dx\leq \epsilon.$$
We can therefore assume w.l.o.g. that $f$ is positive.

*$f\ 1_{[-n,n]}\to f$ pointwise, and by the monotone convergence theorem we can take $n$ such that $$\int_\mathbb{R}|f-f\ 1_{[-n,n]}|dx\leq\frac{\epsilon}{2}.$$
If $h$ is R-simple and $\int_\mathbb{R}|f\ 1_{[-n,n]}-h|dx\leq\frac{\epsilon}{2}$, then w.l.o.g. $h$ is also supported in $[-n,n]$, and we have $$\int_\mathbb{R}|f-h|dx\leq\int_\mathbb{R}|f-f\ 1_{[-n,n]}|dx+\int_\mathbb{R}|f\ 1_{[-n,n]}-h|dx\leq\epsilon.$$
We can therefore assume w.l.o.g. that $f$ is compactly supported.

*The definition of Lebesgue integral affords us a simple function $g$ dominated by $f$ (hence also compactly supported) such that $\int_\mathbb{R}|f-g|dx<\frac{\epsilon}{2}$. Again, if $h$ is R-simple and $\int_\mathbb{R}|g-h|dx\leq\frac{\epsilon}{2}$, then
$$\int_\mathbb{R}|f-h|dx\leq \int_\mathbb{R}|f-g|dx + \int_\mathbb{R}|g-h|dx\leq\epsilon.$$
We can therefore assume w.l.o.g. that $f$ is a compactly supported simple function. Moreover, by a similar process to (1), we can assume w.l.o.g. that $f$ is a compactly supported indicator function.
We're left needing only to prove the proposition for $f=1_A$ for some bounded and measurable $A$. The regularity of the Lebesgue measure admits $K\subset A\subset G$, $K$ compact and $G$ open, such that $\int_{G\setminus K}dx<\epsilon$. By compactness, we have finitely many open intervals $E_1,\ldots,E_n$ such that $K\subset \bigcup E_j\subset G$, and it can now be easily shown that $E^\prime_i = E_i\setminus\bigcup_{j=1}^{i-1}E_j$ is a finite union of disjoint intervals and that $\int_\mathbb{R}|f-\sum 1_{E^\prime_i}|dx\leq\epsilon$.
