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What does the form étant mean in the following sentence, excerpted from an article by E. Szpilrajn:

"La fonction $f_e$ étant mesurable $(K)$, tous les ensembles de la suite $e$ et leur complémentaires appartiennent à $K$."

  1. Does it mean: "$f_e$ being measurable w.r.t. $K$, all the sets of the sequence $e$ and their complements belong to $K$", implying that the measurability of $f_e$ is an established fact at this point.

  2. Or does it mean: "Whenever $f_e$ is measurable w.r.t. $K$, all the sets of the sequence $e$ and their complements belong to $K$", implying a condition, whose premise might not be satisfied, i.e. $f_e$ may not always be measurable w.r.t. $K$, but if it is, all the sets of the sequence $e$ and their complements belong to $K$?

Evan Aad
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    Option 1. Or "Since $f_e$ is measurable w.r.t. $K$ ..." – Martin Apr 13 '13 at 14:48
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    Any intuition I have for French suggests that étant is the present participle of être, but I'm not native French so not certain. Thus, 1. has my vote. – Lord_Farin Apr 13 '13 at 14:49

4 Answers4

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Yes, étant is the present participle of être, the French verb meaning to be. So hypothesis 1. is correct.

Warning: For all questions on the French language always be sure to only trust people with typical French names, like Szpilrajn or yours truly

G. Elencwajg

Did
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Georges Elencwajg
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  • Thank you. I'm not quite convinced, but peer pressure has done me in. – Evan Aad Apr 13 '13 at 15:50
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    @EvanAad This must be the worst reason I can imagine (and a quite disappointing comment to read for those of us who tried to provide facts and arguments in support of their opinions). – Did Apr 13 '13 at 15:56
  • @Did: I'm sorry to disappoint you, but the truth is, it still doesn't make sense to me in the context of the article. I explained myself in the comment below and, while i agree that i did not quote the piece verbatim, the fact remains that $K$ is introduced as an arbitrary family of sets and there is no reason at the point (a) to assume that $f_e$ is measurable with respect to it. The reference to 1.2(iii) doesn't justify the measurability of $f_e$, rather in tandem with the assumption of the measurability of $f_e$, it yields the consequence that $e$'s sets and their complements belong to $K$. – Evan Aad Apr 13 '13 at 16:05
  • @Did: I have to disagree with you and here's a simple counter-example. Let $E$ be some non-trivial subset of $X$ and define $K:={X}$, $E_i:=E$. Then $f_e$ is just $f_E=\mathbb{1}_E$. According to 0.4(i), if $X\in K$, $f_E$ is $(K)$ measurable iff $E\in K$ and $X\backslash E\in K$, which is clearly not the case here. – Evan Aad Apr 13 '13 at 16:40
  • @EvanAad I reread the paper. Please see my answer. – Did Apr 13 '13 at 17:21
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Here is a summary of the situation. We begin by two quotes from the paper:

1.3 Suites d'ensembles appartenant à un $\sigma$-anneau. Soient: $K$ une classe de sous-ensembles d'un espace $X$ et $e=\left\{E_n\right\}$ une suite de sous-ensembles de $X$. Les ensembles $C_n^0$ et $C_n^2$ étant ouverts dans $C$ (voir 1.1(iii)) et l'ensemble $f_e(X)$ étant contenu dans $C$, il résulte de 1.2(iii) que

a) La fonction $f_e$ étant mesurable $(K)$ ${}^{1)}$, tous les ensembles de la suite $e$ et leurs complémentaires appartiennent à $K$.

${}^{1)}$ au sens de 0.4.

0.4 $K$ étant une classe de sous-ensembles d'un espace $X$ et $Y$ un espace métrique, je dis qu'une fonction $f$ définie sur $X$ est $(K)$ mesurable, lorsqu'on a $f^{-1}(G)\in K$ pour chaque sous-ensemble ouvert $G$ de $Y$. (...) Si $X\in K$, on a évidemment :

(i) Pour que la fonction $f_E(x)$ soit mesurable $(K)$, il faut et il suffit qu'on ait : $E\in K$ et $X\setminus E\in K$. (...)

There are two options to understand étant, either it means if or it means since.

In Les ensembles $C_n^0$ et $C_n^2$ étant ouverts dans $C$ (voir 1.1(iii)), étant means since (note that 1.1(iii) states that the sets $C_n^i$ are open and closed in $C$).

In l'ensemble $f_e(X)$ étant contenu dans $C$, étant means since because the inclusion $f_e(X)\subseteq C$ is always true (and is obvious from the definition of $f_e$).

But, in La fonction $f_e$ étant mesurable $(K)$, étant cannot mean since because $f_e$ is not always measurable $(K)$. A simple case when $f_e$ is not measurable $(K)$ is given by the OP in a comment:

Let $E$ be some non-trivial subset of $X$ and define $K:={X}$, $e=(E_n)_n$, $E_n=E$. Then $f_e=\mathbf 1_E$. According to 0.4(i), if $X∈K$, $f_E$ is $(K)$ measurable if and only if $E∈K$ and $X\setminus E∈K$, which is clearly not the case here.

Hence, the only option is that étant in La fonction $f_e$ étant mesurable $(K)$ means if.

To conclude, étant is used alternatively as since and as if and it is used as if in the occurrence the OP mentions (hence, Option #2).

Did
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    Kudos to the OP for having stuck to their guns, and apologies from me for a too cursory first reading of Szpilrajn's paper. I quickly spotted an example of étant meaning since and I deduced (falsely) that this extended to every occurrence of étant in the paper. Sorry again. – Did Apr 13 '13 at 17:20
  • Merci beaucoup! :) – Evan Aad Apr 13 '13 at 17:28
  • De rien. À dire vrai, je ne suis pas très fier d'être passé trop vite sur votre question la première fois. Merci d'avoir insisté. – Did Apr 13 '13 at 17:30
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As in the comments, I'd go for (1). Étant is the present participle of être, so 'being' is a good translation. Not only that, but a common phrase is 'étant donné que', which has the same meaning as 'given that', so it seems like the meaning would carry over to this sentence.

Not a native French speaker though, so may be best to wait until one comes along and answers!

Tim
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I'm gonna sound like a snotty, ignorant jerk, but i have to disagree with everyone, even though i'm not a native French speaker (obviously...), whereas at least one of the respondents is and even though i have the highest regard to all the respondents, especially to Davide Giraudo, who has answered some of my mathematical questions in the past.

Option #1 just doesn't make sense in the context of the article, imo. Here's a larger excerpt surrounding the sentence in question. Please let me know if you agree with me.

Soient: $K$ une classe de sous-ensembles d'un espace $X$ et $e=\left\{E_n\right\}$ une suite de sous-ensembles de $X$. Les ensembles $C_n^0$ et $C_n^2$ étant ouverts dans $C$ et l'ensemble $f_e(X)$ étant contenu dans $C$, il résulte que

a) La fonction $f_e$ étant mesurable $(K)$, tous les ensembles de la suite $e$ et leur complémentaires appartiennent à $K$.

Nous allons démontrer qu'on a d'autre part:

b) $K$ étant un $\sigma$-anneau et $e$ étant une suite d'ensembles appartenant avec leur complémentaires à $K$, la fonction $f_e$ est mesurable $(K)$.

In my opinion the "étant"-s used in the first paragraph are indeed meant to establish facts, but the "étant"-s used in (a) and (b) introduce the legal conditional that can sometimes be found in legal texts (again, i'm hardly a legal expert, more like an outlaw, but it sounds right to me).

The article in question is linked here. The rogue excerpt can be found in the beginning of section 1.3.

Evan Aad
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  • @Did: But $K$ is introduced as just an arbitrary family of sets. It is not given a-priori that $f_e$ is measurable with respect to it. How can we tell in (a) that $f_e$ is measurable with respect to $K$? Rather (a) seems to say: "If $f_e$ is measuragle w.r.t. $K$, then ... (bla bla bla)" – Evan Aad Apr 13 '13 at 15:38
  • After more careful consideration, I must withdraw my first comment. See answer. – Did Apr 13 '13 at 17:28