I've been reading [Paolini-Stepanov arcticle](https://arxiv.org/abs/1303.5664) and in section 4, at page 6, they define a metric current from a transport:
$$T\_{\eta}(\omega)=\int\_{\Theta}[[\theta]](\omega)d\eta(\theta),$$
which satisfies
$$\mu\_{T\_\eta}\leq\int\_{\Theta}\mu\_{[[\theta]]}d\eta.$$
Where $\Theta$ is the set of Lipschitz curves, $\mu\_T$ is the mass of the current and $[[\theta]]$ is the current associated to a Lipschitz curve.


Then they build an example in which this inequality is strict.
In the example we take an $\eta\_1$ concentrated over horizontal segments in $Q=[0,1]^2$ (which define the set $\Theta\_1$) going from left to right and defined on a Borel set $e$ to be the 1-Length of the set of the starting points of the curves in $e$
$$\eta\_1(e):=\mathcal{H}^1(e\_0(e\cap\Theta\_1).$$
Then they build $\eta\_2$ in the same way on the vertical segments ion $Q$.
Now they show that $T\_{\eta\_1+\eta\_2}$ satisfies the inequality above in a strict sense.


The conclusion of the paper is that any acyclic normal current comes from a transport and have equality in that relation.


Since the obtained current cannot not be acyclic or normal, I think I am missing something.
What am I missing?


Addendum: during the definition of this current they represent them with the notation 
$$T\_{\eta\_1}=\bar{e}\_1\wedge\mathcal{L}^2\llcorner Q$$
Which is alien to me. Maybe by understanding it I could see the problem above.



Metric 1-current decomposition
I think for any $\bar e \in \mathbb R^2$ the current $\bar e \wedge \mathcal{L}^2 \llcorner Q$ is defined in their paper via $<\bar e \wedge \mathcal{L}^2 \llcorner Q, \varphi> = \int\_Q \bar e\cdot \varphi \, dx$ for any bounded Borel $\varphi \colon Q \to \mathbb R^2$. Therefore $T\_{\eta\_1} = \bar e\_1 \wedge \mathcal{L}^2 \llcorner Q$, $T\_{\eta\_2} = \bar e\_2 \wedge \mathcal{L}^2 \llcorner Q$ and consequently $T\_{\eta\_1+\eta\_2} = (\bar e\_1 + \bar e\_2) \wedge \mathcal{L}^2 \llcorner Q$. Note that this current actually *is normal* by theorem 4.2, or more explicitly because $\partial (T\_{\eta\_1+\eta\_2}) = s \wedge \mathcal H^1 \llcorner \partial Q$, where $s(x) = \mathrm{sign}(x\_1 + x\_2 - 1)$.


