Writing $f= f_1 + f_2,$ where $f_1\in L^1, \, f_2 \in L^\infty$

Question

Suppose $\int_E |f| < \infty$ for any measurable $E\subset\mathbb{R}^n$ with $m(E) < \infty.$ Prove that there are $f_1 \in L^1(\mathbb{R}^n)$ and $f_2 \in L^{\infty}(\mathbb{R}^n)$ such that $f=f_1+f_2$.

My idea is that we put the "thing" not bounded in $f_1 \in L^1(\mathbb{R}^n)$ and put the "thing" that is bounded but not summable (may be some bounded value at infinity?). But I do not know how to write it rigorously.

Answer 1

WLOG, $f$ is nonnegative. Let $\mu$ be Lebesgue measure on $\mathbb R^n.$ Suppose we can show that

$$\tag 1 \int_{\{f>m\}} f\, d\mu < \infty\,\,\text { for some } m\in \mathbb N.$$

Then we can take $f_1 = f\cdot\chi_{\{f>m\}}, f_2 = f\cdot\chi_{\{f\le m\}}.$

Suppose $(1)$ fails. Then $\int_{\{f>m\}} f\ d\mu = \infty$ for all $m.$ By the hypothesis on $f,$ this implies $\mu(\{f>m\})=\infty$ for all $m.$

Let $m = 1.$ Then $\mu {\{f>1\}} = \infty.$ By the intermediate value property of Lebesgue measure, there exists a set $E_1 \subset \{f>1\}$ with $\mu(E_1) = 1.$ We continue inductively: Suppose disjoint $E_1, \dots, E_m$ have been chosen with $E_k \subset \{f>k\}$ and $\mu(E_k) = 1/k^2,$ $k =1,\dots ,m.$ Because $\mu(\{f>m+1\}) = \infty,$

$$\tag 2 \mu(\{f>m+1\} \setminus (E_1 \cup \cdots \cup E_m)) = \infty$$

as well. Thus, using the IVP again, there exists $E_{m+1}\subset \{f>m+1\} \setminus (E_1 \cup \cdots \cup E_m)$ with $\mu(E{m+1}) = 1/(m+1)^2.$

By induction we thus have a sequence $E_1, E_2, \dots$ with the properties described above. Let $E= \cup_m E_m.$ Then $\mu(E) < \infty.$ But

$$\int_E f\, d\mu = \sum_{m=1}^{\infty} \int_{E_m} f\, d\mu \ge \sum_{m=1}^{\infty} m\cdot\frac{1}{m^2} = \sum_{m=1}^{\infty} \frac{1}{m} = \infty.$$

That contradicts the hypothesis on $f.$ Thus $(1)$ holds, and we're done.

Answer 2

Are you sure, that this is correct? I have a counterexample in mind, which basically consists of repeating integrable poles: $$f_k(x)=\left\{\begin{array}{cc}\frac{1}{\sqrt{x-k}},& k<x<k+1\\ 0,& \mbox{else}\end{array}\right.$$ for $k\in\mathbb{N}$. Now let $f(x)=\sum_{k=1}^\infty f_k(x)$. $f$ has a pole at every natural number but is locally integrable. Now assume we have found $g_1\in L^1(\mathbb{R})$ and $g_2\in L^\infty(\mathbb{R})$ with $f=g_1+g_2$. Hence there exists a $C>0$ such that $|g_2|\leq C$. We also find an $\varepsilon>0$ such that for every $k\in\mathbb{N}$ we have $f(x)>C$ for every $x\in(k,k+\varepsilon)$. Hence $$\int_k^{k+\varepsilon}|g_1|\,dx \geq \int_k^{k+\varepsilon}|f - C|\,dx>\delta>0$$ for some fixed $\delta>0$ independend of $k$. Therefore if we sum over all $k$, we obtain $$\int_\mathbb{R} |g_1|\, dx = \infty.$$ EDIT:

With the information that $f$ needs to be integrable on a set with finite measure the counterexample does not work anymore. Set for example $$E=\cup_{k\in\mathbb{N}}(k,k+\frac{1}{k^2}).$$ Then $|E|=\sum_{k=1}^\infty\frac{1}{k^2}<\infty$ but $\int_0^{\frac{1}{k^2}}\frac{1}{\sqrt{x}}\, dx=2\frac{1}{k}$. Therefore $$\int_E f(x)\, dx=\sum_{k=1}^\infty \frac{1}{k}=\infty$$

Answer 3

I'm really not sure about that, if i take for example in $\mathbb R$, $f=\frac{1}{\sqrt x} \cdot \mathbb 1_{[0,1)}, \forall x \in [0,1)$ and then define it periodically as the translation of that in each interval $[k,k+1), k \in \mathbb Z$, we have $f \in L^1_{loc}$.

Now we want $f-f_2$ to be in $L^1(\mathbb R)$(so we want to choose $f_2$ in order to minimize the $L^1$ norm of $f-f_2$, since $f_2 \in L^{\infty}$ then we also have $f_2(x)\le h, \forall x \in \mathbb R$).

Now suppose that the claim holds true for some $f_1,f_2$ and take $\|f_2\|_{\infty}=h$, we now define for each interval $[k,k+1)$ the sets: $$A_k=\{x \in [k,k+1):f(x)\le h \},B_k=\{x \in [k,k+1):f(x)> h \}$$ (note that since $f$ is periodic the sets have the same length for each interval $[k,k+1)$ and the sets $B_k$ have measure $>0$) then the best that we can hope for is that in each interval $[k,k+1)$ we have: $$f(x)-f_2(x)=0, \forall x \in A_k $$ and $$f(x)-f_2(x)=f(x)-h>0, \forall x \in B_k.$$ Now in order for $f-f_2$ to be in $L^1(\mathbb R)$ it has have finite integral over $\mathbb R$, since the function is periodic we can just see that the integral in each interval is $\int_k^{k+1} f-f_2 \;dx=\int_{B_k} f(x)-h \;dx=a>0$, so the function will have an integral over $\mathbb R$ equal to: $$\int_{\mathbb R} f-f_2 \; dx=\sum_{k=-\infty}^{+\infty} \int_k^{k+1} f-f_2 \;dx=\sum_{k=-\infty}^{+\infty} \int_{B_k} f(x)-h \;dx=\sum_{k=-\infty}^{+\infty} a=+ \infty.$$ That's the reason why i think the claim is false, if you are sure about the claim then we have to wait for some more expert user that can help us.

Answer 4

The proposition is not true in general if you mean $f\in L_{1,loc}(\mathbb{ R})$. Any continuous function $f$ is integrable on any compact subset of $\mathbb{R}$. But if $\lim_{x \to \infty} f = \infty$ it can not be written in this form.