In my readings I often come across terms like "induced measure" or "induced Lebesgue measure". For example:

$$\int_{\mathbb{B}^n}u\frac{\partial v}{\partial x_j}\;dx = \int_{\mathbb{S}^{n-1}}uv\frac{x_j}{|x|}\;d\sigma - \int_{\mathbb{B}^n}v\frac{\partial u}{\partial x_j}\;dx$$ where $d\sigma$ denotes the induced Lebesgue measure on the sphere.

Unfortunately though, I've never really seen anyone give a rigorous definition to this phrase. Sure, in the above example, we understand how to parametrize the $n$ ball, $\mathbb{B}^n$, and the $n-1$ sphere, $\mathbb{S}^{n-1}$, and so the integrals are easy to compute and not very confusing. Sometimes, however, it appears in the context of a general hypersurface in $\mathbb{R}^n$ (say $\Sigma$, with integrals involving $d\Sigma$), or when integrating over a subset (submanifold) of some higher-dimensional space.

In a completely general setting like this, I am at a little bit of a loss when it comes to understanding how these separate "surface" measures or "induced" measures are really defined, when I'm lacking an explicit parametrization.

It also doesn't help that traditional vector calculus classes seem to always fall short of doing anything past 3 dimensions. For example, I've never seen a class teach its students how to carry out a (2-D) surface integral in anything except $\mathbb{R}^3$; so they would have no idea how to go about finding the surface area of, say a Clifford Torus in $\mathbb{R}^4$.

And in general, I've never seen them address a general technique for integrating over an $n$-dimensional body, embedded in $n<m$-dimensional space. It seems like everything they're taught is in terms of the cross product, which fails to be of any use in $\mathbb{R}^{n>3}$.

Now I realize that at least some of the hypersurface examples from calculus, that I've mentioned above, can be addressed using the generalized Stokes' theorem: $\int_{D}\text{d}\omega = \int_{\partial D}\omega$. But it seems to me that when we are getting to the realm of measure theory, higher-level analysis, and Riemannian geometry, the word "induced" is probably thrown around for a good reason. Let me clarify what I mean:

  • If we are on a Riemannian manifold $M$, then its metric $g$ defines a volume element for the the manifold, $dV_g = \sqrt{\det g}\;dV$, where $dV$ is the Euclidean measure for your coordinate patch in $\mathbb{R}^n$. If we then consider an immersed submanifold $N\hookrightarrow M$, we have a rigourous definition for the induced metric that $N$ inherits from $M$.
  • If we have a topological space $(X,\tau)$, then it induces a subspace topology on any subset $Y\subset X$, $\tau_Y = \{Y\cap U : U\in\tau\}$. Similarly for any subset $Y\subset X$ of a $\sigma$-algebra $(X,\Sigma)$, we have the sub-$\sigma$-algebra $(Y,\Sigma_Y), \Sigma_Y = \{Y\cap A : A\in\Sigma\}$.

In these settings, the word "induced" has a very specific meaning, and essentially boils down to the idea of a subset inheriting some sort of property from a larger set it belongs to. So it would make sense that there should be some concrete way of having a measure space induce some kind of measure on a subset of itself.

At first glance, since a measure space $(X,\Sigma,\mu)$ can give rise to a sub-$\sigma$-algebra (as mentioned above) then you might want to simply take the restriction of $\mu$ to these subsets in $\Sigma_Y$. But the problem is that any subset $E$, without full dimension, will of course have measure zero, $\mu(E)=0$. So this would make for a lousy way to define integration over this subset.

So maybe we only concern ourselves with Riemannian manifolds, and just define the induced measure as the one that arrises from the induced metric of some parent manifold. But this seems limiting for a couple reasons:

  1. Integration is now limited to Riemannian manifolds, and no longer over a general measure space. For example, while integration with respect to the counting measure makes sense over $\mathbb{N}$, I can't see any way to interpret this as integration over a Riemannian manifold; and so the question of how the counting measure "induces" another measure would be meaningless.
  2. If we consider a (smooth) subset $S\subset\mathbb{R}^n$, we can build a measure space on $S$ from scratch, similar to how we build the Lebesgue measure on $\mathbb{R}^n$ in measure theory. My question then is: are we always able to realize integration (with respect to this measure) over $S$ as integration with respect to some induced metric, by considering $S\subset M$, for some manifold $(M,g)$?

To me, a measure (and anything that it induces) should be in the spirit of measurable sets, and not be restricted to differential manifolds. The Lebesgue measure $\mu$ in $\mathbb{R}^3$ gives us the volume of the unit ball as $\frac{4}{3}\pi$; shouldn't $\mu$ induce a measure on $\mathbb{S}^2$, to give us $4\pi$ surface area? And whatever method we choose, shouldn't it apply to abstract measures as well, not just Lebesgue measure?

  • 1
    $\begingroup$ After checking this question in physics, I wanted to find a mathematical reference for such a question. I have found one in "A course on integration, Nicolas Lerner" (2014), §5.5 p.238. This is another approach. $\endgroup$
    – Noix07
    Oct 12, 2018 at 12:51
  • $\begingroup$ Thank you. That's an interesting reference. $\endgroup$
    – Patch
    Oct 12, 2018 at 14:51
  • $\begingroup$ Another interesting reference: "Éléments de distributions et d’équations aux dérivées partielles" (Dunodn 2002) by Claude Zuily, (2.11), (2.16) p.42 'till p.44: the measure on the boundary of a domain $\omega$ is obtained as the weak derivative of the characteristic function $\chi_{\Omega}$. cf. also (2.19) p.45, when the boundary is given as the graph of a function $\endgroup$
    – Noix07
    Apr 11, 2022 at 19:17

1 Answer 1


The word "induced" does not always refer to "a subset inheriting some sort of property from a larger set it belongs to". It also refers to one kind of structure defining another kind of structure on the same set. Examples:

  • norm induced by an inner product
  • metric induced by a norm
  • topology induced by a metric

All of these can refer to the same underlying set, e.g. $\mathbb{R}^n$.

In the context of your quote at the beginning, I would interpret "induced" as in

measure induced by Riemannian metric [by means of the volume form].

Here is a usage example from MathOverflow.

But if you want to induce the measure on $S^2$ coming from the ambient space $\mathbb{R}^3$, I can suggest some approaches:

  1. Consider the Hausdorff measure $\mathcal H^2$ on $\mathbb{R}^3$. Its restriction to $S^2$ is the surface measure.

  2. Given $A\subset S^2$, define the [upper] Minkowski content of $A$ as $$m^*(A) = \limsup_{r\to 0} \frac{\mu(A_r)}{2r}\quad \text{where } A_r=\{x\in\mathbb{R}^3 : \operatorname{dist} (x,A)\le r\} $$ For nice sets this is the same as surface area; unfortunately for uglier sets countable additivity fails. Also, the $\limsup$ and $\liminf$ are different in general.

  3. Fix the subadditivity issue in #2 by defining the [upper] packing measure of $A$ as $$\inf\left\{ \sum_{j\in\mathbb{N}} m^*(A_j) : \bigcup_{j\in\mathbb{N}} A_j = A\right\}$$ This is an ugly definition, but it has countable sub-additivity built-in. Since the resulting measure is invariant under isometries of $S^2$, it is in fact the surface area.

  • $\begingroup$ Thanks. The Minkowski content definition is along the lines of what I was looking for; it's also good to know that this definition has some important limitations. The stuff in the Overflow link was also very helpful. Thank you. $\endgroup$
    – Patch
    Mar 18, 2015 at 21:36
  • $\begingroup$ In this answer, $\mu$ refers to the Lebesgue measure in $\mathbb{R}^3$, right? $\endgroup$
    – kdbanman
    Apr 10, 2020 at 17:36

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