# A problem on Lebesgue dominated convergence theorem

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

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.

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.

• Can I define $g= -\delta$ on A where A = {$t: f(t)<−δ$} and since g is a simple function $\int_A g <0$ and $\int_A f < \int_A g$? and that is a contradiction? – user68099 May 29 '14 at 5:48
• @user68099 Yes, because $$0\leq\int_A f \leq -\delta\cdot|A|$$ so $|A|$ have measure 0. Also note that we only have $$\int_A f \leq \int_A g$$ – AD. May 29 '14 at 5:55

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.