Find $\delta >0$ such that $\int_E |f| d\mu < \epsilon$ whenever $\mu(E)<\delta$

Question

I am studying for a qualifying exam, and I am struggling with this problem since $f$ is not necessarily integrable.

Let $(X,\Sigma, \mu)$ be a measure space and let $$\mathcal{L}(\mu) = \{ \text{ measurable } f \quad| \quad \chi_Ef \in L^1(\mu) \text{ whenever } \mu(E)<\infty\}.$$ Show that for any $f\in \mathcal{L}(\mu)$ and any $\epsilon >0$ there is $\delta >0$ such that $\int_E|f| d\mu < \epsilon$ whenever $\mu(E)< \delta$.

A technique I've used in other similar problems is to define $A_n = \{ x\in X \, | \, 1/n \leq |f(x)| \leq n \}$ and let $A = \displaystyle \bigcup_{n=1}^\infty A_n$. We can also define $A_0 = \{ x\in X \,|\, f(x) = 0\}$ and $A_\infty= \{x\in X\, | \, |f(x)| = \infty\}$. The part where I'm stuck is now that $$\int_X|f|d\mu = \int_{A_0} |f|d\mu + \int_{A} |f| d\mu + \int_{A_\infty} |f|d\mu$$ where the first term on the right is zero, and I want the last term on the right to be zero.

Is there another way to go about this problem? Explanations are helpful to me since I'm studying and I don't want to confuse myself further. Thanks!

Answer 1

Suppose not: there is $\varepsilon_0\gt 0$ such that for each positive $\delta$, there is a measurable set $A$ such that $$\mu(A)\lt \delta\quad \mbox{ and }\quad \int_A|f|\mathrm d\mu\gt\varepsilon_0.$$ In particular, for each integer $k$ and $\delta:=2^{-k}$, there exists $A_k$ of measure smaller than $2^{-k}$ for which $\int_{A_k}|f|\mathrm d\mu\gt\varepsilon$. Define $A:=\bigcup_k A_k$ (a set of finite measure). We have $$\varepsilon_0\lt \int_{A_k}|f|\mathrm d\mu\leqslant n\mu(A_k)+\int \chi_{A_k}\chi_{\{|f|\gt n\}}|f|\mathrm d\mu\\\leqslant n2^{-k}+\int \chi_A\chi_{\{|f|\gt n\}}|f|\mathrm d\mu,$$ hence $$\varepsilon_0\leqslant\int \chi_A\chi_{\{|f|\gt n\}}|f|\mathrm d\mu,$$ and by monotone convergence, $\lim_{n\to\infty}\int \chi_A\chi_{\{|f|\gt n\}}|f|\mathrm d\mu=0$, a contradiction.

Answer 2

My try: the best thing to do in my opinion is working by absurd. Suppose the thesis does not hold.

Then $\exists \varepsilon >0$ such that $\forall n \geq 1$ exists $E_n \in \Sigma$ such that $\mu (E_n) < 2^{-n} \varepsilon$ BUT $\int_{E_n} |f| d \mu > \varepsilon$.

Then call $E=\bigcup_{n\geq1} E_n$. We have that $\mu(E) < \varepsilon$ but $$\int_{E} |f| d\mu = \sum_{n\geq1} \varepsilon = +\infty$$

This contradicts the assumption on $f$.

The only problem is that this works only if $\{E_n \}_{n\geq1}$ are pairwise disjoint.

Answer 3

EDIT: There is a flaw in this proof for $X \neq R^n$ (which is probably really something like $X$ not hausdorff) as pointed out in the comments.

Here is a proof that follows the train of thought I had where I initially guessed the statement given was wrong. The ideas here are similar to Crostul's answer.

First, we must have the measure of $E_n = f^{-1}([2^n, \infty))$ eventually be finite. That is, there is an $N$ such that $n \geq N$ implies $\mu(E_n) \leq m(E_N) < \infty$.

Suppose this wasn't true. That is, $\mu(E_n) = \infty$ for all $n$. Then for each $E_n$ there is an infinite amount of room to inductive pick a set $A_n \subset E_n \setminus \cup_{k=1}^{n-1} A_k$ with $2^{-n-1} \leq \lambda(A_n) \leq 2^{-n}$ since $\mu(E_n \setminus \cup_{k=1}^{n-1} A_k) = \infty$. But this means $A = \cup_{k=1}^\infty A_k$ is finite and yields the contradiction

$$\infty = \sum_{k=1}^\infty \frac{1}{2} \leq \sum_{k=1}^\infty \mu(A_k) 2^k \leq \sum_{k=1}^\infty \int_{A_k} f d\mu = \int_A f(x) dx < \infty.$$

So next let $E_N = f^{-1}([2^N,\infty)$ and note that $\chi_{E_N} f$ is an $L^1$ function and $L^1$ functions have the desired property (try to prove this if you don't know it). So it remains to show that the property holds for $\chi_{E_N^c} f$.

This is easy since $\mu(E)<\delta$ with means $$\int_E \chi_{E_N^c} f d \mu \leq \mu(E \cap E_N^c) 2^N \leq \mu(E) 2^Nv\leq \delta 2^N$$ because $f$ is bounded by $2^N$ on $E_N^c$. So we pick $\delta = 2^{-N} \epsilon$.

Finally, all that remains is to show that if the desired property holds for $g$ and $h$, then it holds for $g+h$ and smooth out a couple rough edges in this argument, but that is easy to do.