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 $\|f-\phi\|_{L^1}<\epsilon$. enter image description here

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 $\|f-\phi\|_{L^1(E)}\to 0$

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

  • $\begingroup$ It is doubtful if the approximation result you have used is permitted here. $\endgroup$ – Kavi Rama Murthy Oct 30 '19 at 0:01
  • $\begingroup$ what is this theorem? and what is $\triangle$ operator? $\endgroup$ – domath Oct 30 '19 at 0:01
  • $\begingroup$ is the second solution I wrote working? $\endgroup$ – domath Oct 30 '19 at 0:03
  • $\begingroup$ @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 $\endgroup$ – Marios Gretsas Oct 30 '19 at 0:09
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    $\begingroup$ @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 $\endgroup$ – Marios Gretsas Oct 30 '19 at 1:02

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