$\forall \epsilon>0 , \exists$ a finite linear combination of charactristic functions of intervals such that $\|f-\phi\|_{L^1}<\epsilon$.

Let $$f$$ be non-negative function in $$L^1=L^1([0,1],\lambda)$$. Prove that for each $$\epsilon>0$$ there exists a finite linear combination of charactristic functions of intervals , $$\phi = \sum c_i \chi_{[a_i,b_i)}$$ , such that $$\|f-\phi\|_{L^1}<\epsilon$$.

My understanding of the question is that it is asking to prove that there exists a sequence of step functions that converges in $$L^1$$ to function $$f$$. Is it right?

($$\textbf{Side question}$$: Is the question as of showing that we can approximate any measurable function $$f$$ by a sequence of simple functions? If not what is the difference between this question and that?)

My attempt:

Let $$E=[0,1]$$ since $$f\in L^1$$ and $$\lambda(E)<\infty$$, we have that $$f\leq L$$ for some $$L>0$$. let $$E=\bigcup_{i=1}^M E_i$$ , $$E_j\cap E_l=\phi , \forall j\neq l$$ and $$h=\sum a_i \chi_{E_i}$$ be a sequence of simple functions that converges pointwise to $$f$$ such that $$|f-h|<\frac{\epsilon}{2}$$.

let $$A^i = \{x | \chi_{E_i}\neq \chi_{\cup[a_i,b_i)}\}$$, then let the $$\lambda(A^i)<\frac{\epsilon}{2Ma_i}$$ . so $$\|h-\phi\|_{L^1(E_i)}<\frac{\epsilon}{2M}$$.

so we have \begin{align} \|f-\phi\|_1 & \leq \|f-h\|_1+\|h-\phi\|_1 = \|f-h\|_1 + \int_{\cup E_i} |f-\phi|d\lambda \\ & = \|f-h\|_1 + \sum_{i=1}^M\int_{E_i} |f-\phi|d\lambda\\ & = \|f-h\|_1 + \sum_{i=1}^M\|h-\phi\|_{L^1(E_i)}\\ & \leq \frac{\epsilon}{2} + M.\frac{\epsilon}{2M}\\ & = \epsilon \end{align}

$$\textbf{Question}:$$My friend was also suggesting that I could not see what is the problem with that if it is not working , the solution looked simple though. Taking an increasing sequence of simple function $$\phi_n$$ that converges pointwise to $$f$$ . That is $$\phi_n \leq f$$. so we'd have that using monotone convergence theorem , $$\lim \int_E\phi_n =\int_E \phi \leq \int_E f$$ then it follows that also $$\|f-\phi\|_{L^1(E)}\to 0$$

• An integrable function need not be bounded. – Kavi Rama Murthy Oct 29 '19 at 23:50
• but if the measure of the domain is also finite it is bounded right? – domath Oct 29 '19 at 23:57
• No. $\frac 1 {\sqrt x}$ is integrable on $(0,1)$ but it is not bounded. – Kavi Rama Murthy Oct 29 '19 at 23:59

Why the sets $$A_i$$ have measure $$<\frac{\epsilon}{2Ma_i}$$?.This may not be the case.

Also $$f$$ is not bounded.

You know that simple functions are dense on $$L^1$$ thus you exists $$h=\sum_{i=1}^{k}a_i1_{E_i}$$

So you have to approximate each $$E_i$$.

Let $$\epsilon>0$$

We have that $$\lambda(E_i)<\infty$$ and that $$E_i$$ is measurable.

Thus exist disjoint open intervals $$I_1,...I_{m_{i}}$$ such that $$\lambda(E_i \cap \triangle \bigcup_{j=1}^{m_i} I_j)<\frac{\epsilon}{\sum_{i=1}^k|a_i|}$$

Thus $$\int_0^1|1_{E_i}-1_{\bigcup_{j=1}^{m_i}}| \leq \lambda(E_i \cap \triangle \bigcup_{j=1}^{m_i} I_j)<\frac{\epsilon}{\sum_{i=1}^k|a_i|}$$

Note that the intervals are disjoint so the indicator of the union is the sum of indicators.

Can you continue from here to prove the approximation of $$h$$ by a simple function with intervals?

• It is doubtful if the approximation result you have used is permitted here. – Kavi Rama Murthy Oct 30 '19 at 0:01
• what is this theorem? and what is $\triangle$ operator? – domath Oct 30 '19 at 0:01
• is the second solution I wrote working? – domath Oct 30 '19 at 0:03
• @stat_yale the triangle is the symmetric difference of sets...the second solution with lusin does not work since lusin's theorem involves continuous functions..here you do not need continuous functions – Marios Gretsas Oct 30 '19 at 0:09
• @stat_yale yes..but unfortunately this is not always the case.the class of simple functions with intervals(they are called step functions b.t.w) is a proper subset of the class of simple functions – Marios Gretsas Oct 30 '19 at 1:02