Showing $\lim_{n \to \infty} m(E_n) = 0$, assuming $f > 0$ a.e. and $\lim_{n \to \infty} \int_{E_n}f \,dm =0$

measure-theory

Here is a direct proof: Let $k,n \in \mathbb{N}$. Then

$$\int_{E_n} f \, dm \geq \int_{E_n \cap \left[f>\frac{1}{k}\right]} f \, dm \geq \frac{1}{k} \cdot m \left( E_n \cap \left[f> \frac{1}{k} \right] \right) \geq 0$$

Since $\int_{E_n} f \, dm \to 0$ as $n \to \infty$ we obtain

$$m \left( E_n \cap \left[f> \frac{1}{k} \right] \right) \to 0 \qquad (n \to \infty)$$

for all $k \in \mathbb{N}$. Thus

$$m(E_n) \leq m \left( E_n \cap \left[f> \frac{1}{k} \right] \right) + m \left( \left[f > \frac{1}{k} \right]^c \right) \to m \left( \left[f > \frac{1}{k} \right]^c \right) \qquad (n \to \infty)$$

We have

$$m \left( \left[f > \frac{1}{k} \right]^c \right) \to 0 \qquad (k \to \infty)$$

since $f>0$ a.s. and therefore conclude $m(E_n) \to 0$ as $n \to \infty$.


Hint: given $\epsilon > 0$, write $m(E_n) = m(E_n \cap \{f \ge \epsilon\}) + m(E_n \cap \{f < \epsilon\})$. Use Markov's inequality on the first term.

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