I'm reading Protter and sometimes he says "locally square integrable martingale", and sometimes he says "locally square integrable local martingale", and I wonder if these two are the same.
Protter's definition is a property holds locally for $X$ if there are $T_n$ increasing to $\infty$ such that $X^{T_n}1_{T_n>0}$ has the property.
Then a local martingale is an adapted cadlag $X$ which is locally a martingale.
Protter also proves that a cadlag $X$ which is locally a local martingale is a local martingale.
Now I try to parse "locally square integrable local martingale". If $X$ is such that it is locally a square integrable local martingale, in particular $X$ is locally a local martingale, and hence is a local martingale. Thus the question reduces to is "(locally square integrable) + (local martginale) = locally (square integrable martingale)"?
I think they are the same because if $T_n$ makes $X^{T_n}1_{T_n>0}$ a martingale, and $S_n$ makes $X^{S_n}1_{S_n>0}$ square integrable, then $T_n \wedge S_n$ should make $X^{T_n \wedge S_n}1_{T_n \wedge S_n > 0}$ a square integrable martingale. Is this line of reasoning correct or am I missing something?