As the headline suggests, I'm wondering whether a local continuous martingale is actually necessarily square integrable.
In the text I'm reading this is only mentioned without explanation (the answer may be trivial but I don't see it). I know that the quadratic variation of a continuous local martingale is continuous, does this somehow with Doob-Meyer's decomposition yield the answer?
If you are asking whether a continuous local martingale is square integrable process the answer is NO, as the counterexample of @Jochen demonstrated.
But if you are asking whether a continuous local martingale is locally a square integrable martingale (meaning both properties should hold by the same localizing sequence), the answer is YES. This can be seen by the property that every continuous process locally bounded. By taking the minimum of both sequences (the one for the martingale-property and the one for the boundedness), both properties remain. So you have a localizing sequence which makes your process a bounded (and thus square integrable) martingale.