Confusion about conservation of angular momentum tensor in classical field theory?

In my lectures, we considered the conserved stress energy tensor $$T^{\mu \nu}$$ and noted that we could always add a conserved tensor to it such that $$T^{\mu \nu}$$ is symmetric.

As a consequence, a tensor of higher rank, $$M^{\lambda \mu \nu}= x^\mu T^{\lambda \nu} - x^{\nu}T^{\lambda \nu}$$ is always conserved in that $$\partial _{\lambda} M^{\lambda \mu \nu}=0$$.

What I don't understand is that the claim is that the conservation of $$M^{\lambda \mu \nu}$$ results in the conservation of the total angular momentum tensor:

$$J^{\mu \nu} = \int d^3x M^{0 \mu \nu}$$

In essence, it is claimed that symmetry of stress energy tensor -> conservation of $$M^{\lambda \mu \nu}$$ -> Conservation of $$J^{\mu \nu}$$.

I don't see how this last step holds. I have tried the algebra, but I don't see how you can deduce anything about $$\partial _{\mu} M^{0 \mu \nu}$$, which you would need to vanish for $$\partial _{\mu} J^{\mu \nu}$$ to vanish i.e. for the angular momentum tensor to be conserved, based on $$\partial _{\lambda} M^{\lambda \mu \nu}=0$$ where $$\lambda$$ has not been set to 0, and is a sum over $$\lambda$$.

The quantity $$J^{\mu\nu}(t)$$ isn't a conserved current, it's a conserved quantity. Unlike $$M^{\lambda \mu\nu}(\mathbf{x}, t)$$, it doesn't have spatial dependence; at each time it is a tensor rather than a tensor field. The statement is that it doesn't depend on time at all.
The proof of this statement is just the same as the proof for a rank one tensor, since the extra indices just come "along for the ride". If we know $$\partial_\mu J^\mu(\mathbf{x}, t) = 0$$, then we define $$Q(t) = \int J^0(\mathbf{x}, t) \, d^3x.$$ Then $$Q(t)$$ is conserved because $$\frac{dQ}{dt} = \int \partial_0 J^0(\mathbf{x}, t) \, d^3x = - \int \nabla \cdot \mathbf{J} \, d^3x = - \int \mathbf{J} \cdot d\mathbf{S} = 0$$ where the last integral is at spatial infinity, and we assume $$\mathbf{J}$$ vanishes there. The same proof works for $$M^{\lambda \mu \nu}$$ since the extra two indices don't interfere. (For the case of curved spacetime, see here.)