# showing $\int f_n^+\to \int f^+$

Supose $(f_n)$ be such that $\int|f_n-f|\to 0$, where $(f_n)$ is Lebesgue integrable. Show that $\int_E f_n \to \int_E f$ for all Lebesgue measurable sets $E$, and furthermore that $\int f_n^+\to \int f^+$.

(here $f^+$ denotes $f \vee0$)

Not sure where to start on this. Any hints to begin would be helped. thanks.

## 1 Answer

As $\int |f_n - f| \to 0$, we have $\int_E |f_n - f| \to 0$ for any measurable $E$. Now:

$$\left| \left| \int_E f_n \right| - \left| \int_E f \right| \right| \le \int_E| f_n - f|$$

So the LHS converges to $0$, i.e. $\int_E f_n \to \int_E f$.

Now we also have:

$$\int ||f_n| - |f|| \le \int |f_n - f|$$

So the LHS converges to $0$, hence by above, $\int |f_n| \to \int |f|$.

Therefore, $$2 \int f_n ^+ = \int f_n + \int |f_n| \to 2 \int f ^+ = \int f + \int |f|$$

i.e. $\int f_n ^+ \to \int f^+$.