$\liminf \left( \int_E \phi - \int_E f_k \right) = \int_E \phi - \limsup \int_E f_k$

Question

Wikipedia states that $$\liminf (a_n + b_n) \ge \liminf a_n + \liminf b_n.$$

In the proof of Lebesgue Dominated Convergence Theorem, we have this equation

$$\liminf \left( \int_E \phi - \int_E f_k \right) = \int_E \phi - \limsup \int_E f_k,$$

where $f_k \le \phi$ a.e. and $\phi \in L^1$.

Why do we have the equality in the proof of the Lebesgue Dominated Convergence Theorem?

Answer 1

Maybe it's worth doing this from scratch. Suppose $\underset{n\to \infty}\lim b_n=b.$

Fix $k$ for the moment and note that $\underset{n\ge k}\inf a_n+\underset{n\ge k}\inf b_n\le a_i+b_i$ for all $i\ge k$ so $\underset{n\ge k}\inf a_n+\underset{n\ge k}\inf b_n\le \underset{n\ge k}\inf (a_n+b_n).$ Now, let $k\to \infty$ to get one direction of the inequality.

On the other hand,

$\underset{n\ge k}\lim \inf a_n\ge \underset{n\ge k}\liminf (a_n+b_n)+\underset{n\ge k}\liminf (-b_n)=\underset{n\ge k}\liminf (a_n+b_n)-b$, so subtracting $b$ from both sides gives the other direction of the inequality.

Answer 2

$\lim \inf (a-b_n)=a+\lim \inf (-b_n) =a-\lim \sup b_n$. In the special case when $(a_n)$ is a constant sequence equality holds. .

Answer 3

If $a_n = a$ is constant, then we have equality $$\liminf (a+b_n) = a+\liminf b_n$$

This follows directly from $\inf(a + B) = a + \inf(B)$ where $B$ is the set of subsequential limits of $(b_n)$.