Your intuition is right, but for a simple reason. Note that, since $\mu$ is invariant under $T$, we have
\begin{equation}
\int f\circ T\, \mathrm{d}\mu = \int f\, \mathrm{d}(T\mu) = \int f\, \mathrm{d}\mu \;,
\end{equation}
hence,
\begin{equation}
\int (f\circ T - f)\, \mathrm{d}\mu = 0 \;.
\tag{*}
\end{equation}
Now, if the integrand in (*) is almost surely non-negative, it cannot be non-zero on a set of positive measure.
When you pour a glop of syrup into your oatmeal, the contents of the bowl can be modelled as a set $X$ broken into a disjoint union of two measurable subsets $X = S \sqcup O$: $S$ represents the glop of syrup; and $O$ represents the original oatmeal.
Letting $\chi_S,\chi_O : X \to \{0,1\}$ be the characteristic functions, we have $\chi_S(x)+\chi_O(x)=1$ for all $x \in X$.
The volume measure on $X$ can be modelled as Lebesgue measure $\mu$ restricted to $X$, and normalized so that $\mu(X)=1$. You could then express both the syrup and the oatmeal as densities $\chi_S d\mu$, $\chi_O d\mu$.
The transformation itself is being modelled (somewhat unrealistically) as a measure preserving bijection $T : X \to X$.
I do not know what you mean by $f$, it is not otherwise mentioned in your post. But what I'll say is that $\mu$ itself is invariant under the transformation $T$, and so $\mu(A) = \mu(T(A))$ for all $A$, which can be written as $\int_A d\mu = \int_{T(A)} d\mu$.
One key point of ergodicity, as applied to this "syrup/oatmeal" picture, is that neither $S$ nor $O$ is invariant under $T$: neither $\mu(T^{-1}(S) \Delta S)$ nor $\mu(T^{-1}(O) \Delta O)$ is zero. To put it another way neither of the two characteristic functions $\chi_S$, $\chi_O$ is invariant under $T$.
Nonetheless, as one iterates $T$ more and more (as one stirs the oatmeal more and more), the real key point is that function $\chi_S \circ T^{-n}$ has a limit in some appropriate sense, that limit is $T$-invariant, and that limit is actually the constant function $\mu(S)$ with associated distribution $\mu(S) d\mu$ ($T$ will "distribute the syrup evenly throughout"). For this to work one has to take the limit of the function sequence $\chi_S \circ T^{-n}$ very carefully, usually in $L^2(X)$ or $L^1(X)$.
By the way, I have written my answer under the assumption that $T$ is a bijection, which allows me to be careless and write things like $\chi_S \circ T^{-n}$. One can certainly be more careful and rewrite all of this in the general case that $T$ is not a bijection.
Best Answer
Enciclopedia of mathematics says the following:
"A is called a null (or negligible) set if $\mu_∗(A)=0$; in this case the complement $X\setminus A$ is called a set of full measure (or conegligible), and one says that $x\notin A$ for almost all $x$ (in other words, almost everywhere). Two sets $A,B\subset X$ are almost equal (or equal mod $0$) if $(x\in A) \Leftrightarrow (x\in B)$ for almost all $x$ (in other words, $A\setminus B$ and $B∖A$ are negligible). Two functions $f,g:X \longrightarrow Y$ are almost equal (or equal mod 0, or equivalent) if they are equal almost everywhere. "
https://www.encyclopediaofmath.org/index.php/Measure_space