A problem on Lebesgue dominated convergence theorem

Question

I have the following 2 problems for homework, and I couldn't do the 1st one and need to check if my solution is correct for the 2nd one thanks

1) If $f$ is an integrable function on $ \Bbb R $ such that $ \int_{E} f \ge 0 $ for each measurable set E, prove that $f \ge 0 $ almost everywhere.

2) If {$E_n$} is an ascending sequence of measurable sets and $ E= \bigcup_{n=0}^\infty E_n $ prove that $\lim_{n\to \infty}$ $ \int_{E_n} f= \int_E f $ and state and prove an analogous result for decreasing sequences.

for the 2nd one I used $f_n = f \chi_{E_n} $ and used that $\lim_{n\to \infty} f_n = f \chi_E $ and that $\int_{E_n} f= \int_E f= \int f\chi_{E_n} $ and used the LDCT, and for descending sets used E as the intersection and the same result.

Are they correct?

Any hints for the 1st part is appreciated

Answer 1

For the first one, consider the sets $E_n=\{x\in X: f(x)<-\frac{1}{n}\}$ (where $X$ is the underlying space). What would the hypothesis imply for this sets? Note that $\bigcup_{n=1}^\infty E_n=\{x\in X:f(x)<0\}$.

Your solution for the second seems good to me.

Answer 2

Your argument for 2 is fine if you are assuming that $f$ is non-negative and integrable.

For 1, if it were not true that $f\geq0$ a.e., then there exists $\delta> 0$ such that $\{t:\ f(t)<-\delta\} $ has positive measure. Use it to obtain a contradiction.

Answer 3

Hint for 2

First some assumption on $f$ is missing.

• A counter-example is $E_n=[-n,n]$ and $f(x)=x$ if we just assume $f$ is measurable.

• If we assume $f$ is integrable. Write $g_n = \chi_{E_n} \cdot f$, where $\chi_{E_n}$ denotes the characteristic function of $E_n$. Now choose a standard limit theorem for Lebesgue integrals.