It turns out that convergence in Lp implies that the norms converge. Conversely, a.e. convergence and the fact that norms converge implies Lp convergence. Amazing!

**Relationship between convergence and a.e. convergence:**

Let , . If , then . Conversely, if a.e.\ and , , then . Note that the converse may fail for .

**Proof:**

Assume .

(Case: ).

**Lemma 1:**

If , for all .

**Proof of Lemma 1:**

Hence .

**End Proof of Lemma 1.**

Hence, using and we see that

Thus

Hence .

(Case: .)

By Minkowski’s inequality, and so that as . Done.

**Converse:**

Assume a.e.\ and , .

**Lemma 2:**

For , for .

**Proof of Lemma 2:**

By convexity of for ,

Multiplying throughout by gives

Thus together with Lemma 1, for we have with .

Note that a.e.\ and a.e.\ which is integrable. Also, since . By Generalized Lebesgue’s DCT, we have thus

**(Show that the converse may fail for ):**

Consider . Then a.e.\ where , and . However .