# Symmetry modulo total derivative term in Noether's Theorem

I came across the proof of Noether's Theorem in David Tong's notes (page 14) on QFT. He writes something like,

We say that the transformation $$\delta\phi(x) = \chi (\phi) \tag{1.34}$$ is a symmetry if the Lagrangian changes by a total derivative $$\delta \mathcal{L} = \partial_{\mu}F^{\mu}\tag{1.35}$$ for some set of functions $$F^\mu(\phi)$$.

What is the intuition behind calling that a symmetry? (I do follow the math of how we get a current vanishing after we take its derivative)

Recall that on-shell the Lagrangian density $${\cal L}$$ changes by a total divergence term$$^1$$ $$\delta{\cal L}~\approx~d_{\mu} j^{\mu},\tag{A}$$ where $$j^{\mu}~:=~\frac{\partial {\cal L}}{\partial (\partial_{\mu}\phi)}\delta_0\phi+{\cal L}\delta x^{\mu}\tag{B}$$ is the bare Noether current. The intuition is that on-shell there cannot be any bulk term on the RHS of eq. (A), only a boundary term. Therefore if the infinitesimal transformation $$\delta$$ is only a quasi-symmetry [rather than a strict symmetry] of the Lagrangian density [that is: Tong's off-shell eq. (1.35) is satisfied], then we can still obtain an on-shell continuity equation \begin{align} d_{\mu} J^{\mu}~\stackrel{(D)}{=}~~~&d_{\mu} (j^{\mu}- F^{\mu})\cr ~=~~~&d_{\mu} j^{\mu}-d_{\mu} F^{\mu}\cr ~\stackrel{(A)+(1.35)}{\approx}&~\delta{\cal L}-\delta{\cal L}\cr ~=~~~&0\end{align} \tag{C} by defining the following full Noether current $$J^{\mu}~:=~j^{\mu}-F^{\mu}.\tag{D}$$ In other words, we can always compensate for not having a strict symmetry.
The conserved full Noether charge is then defined as $$Q(t)~:=~\int_V\! d^3x~ J^0(x,t) \tag{E}$$ by a standard argument.
$$^1$$ The $$\approx$$ sign means here equality modulo the Euler-Lagrange (EL) equations, i.e. on-shell.