Calculating a simple integral using abstract measure theory

Question

Let $(X,\mu)$ be a measure space.

For a measurable function $\iota:X\to X$, we have for all measurable $E\subset X$,

$$\int_{\iota^{-1}(E)}g\circ\iota \, d\mu = \int_{E}g \, d(\iota^{*}\mu),$$ where by definition $\iota^{*}(\mu)[A] = \mu(\iota(A))$.

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I'm trying to reconcile this with the more elementary process I used to refer to as "$u$ substitution".

Consider the case of $X = \mathbb{R}$ and $\mu$ the Lebesgue measure, and the integral $\int_{[0,2]}f \, d\mu$, where $f(x) = (x^3)^2 \cdot 3x^2$.

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In the notation of Calculus II, I might done something like:

Let $u = x^3$. Then $du = 3x^2 \, dx$ and thus

\begin{eqnarray*} \int_0^2 f(x) \, dx &=& \int_0^2 (x^3)^2 3x^2 \, dx\\ &=& \int_0^8 u \, du \end{eqnarray*}

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I am trying to do the same transformation using the more abstract notation. Setting $\iota(x) = x^3$, then $g(x) = x^2$, I get

\begin{eqnarray*} \int_{[0,2]}f(x) \, d\mu(x) &=& \int_{\iota^{-1}[0,8]}g(\iota(x))3x^2 \, d\mu(x)\\ &=& \int_{\iota^{-1}[0,8]}g(\iota(x)) \, d\lambda(x)\text{, where }\lambda(E) := \int_{E}3x^2 \, d\mu(x)\\ &=& \int_{[0,8]}g(y) \, d[\iota^{*}\lambda](y)\text{, using the law above} \end{eqnarray*}

which seems to me to make sense only if $\iota^{*}(\lambda) = \mu$. At this point I feel like I'm losing grip on things. How can I prove that

$$\mu(E) = \int_{\iota(E)}3x^2 \, d\mu(x)?$$

I understand this is probably not how these calculations are performed in practice but I'm just trying to convince myself that the law stated above truly is a generalization of the one I learned in my first year of studies.