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I am reading a paper that uses the function $\mathbb{I}(x=0)$. I assume this is a common function, but I haven't seen it before.

From the context it seems like it returns $0$ when the argument is true and $1$ when it is false, but I suspect that it is the opposite and that there is a typo in the paper. Just want to make sure.

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  • $\begingroup$ In general, when $A$ is a subset of $X$, the function $\mathbb{1}_A$ refers to the function that is equal to $1$ on $A$ and $0$ on $X\setminus A$. In case $A = \{a\}$ is a singleton, then we usually denote it $\mathbb{1}_{x=a}$ if $x$ is the variable. Sometimes, in $\mathbb{R}^2$, one can define $\mathbb{1}_{x=y}$ as the function that is equal to $1$ on the diagonal $\Delta = \{(x,x)\in \mathbb{R}^2,~|~ x\in \mathbb{R} \}$, etc. This closely depends on the context. $\endgroup$ – DIdier_ Nov 25 '20 at 18:06
  • $\begingroup$ It is the indicator function notation, which is defined as $$ \mathbf{1}_{A}(x)=\begin{cases}1,&x\in A\\0,&x\notin A.\end{cases}$$ The notation varies depending on the literature, and what you see is one such example. For instance, the notation for the indicator function for the event $\{X=a\}$ includes (but is not restricted to) $$\mathbf{1}_{\{X=a\}},\qquad \mathbf{1}[X=a],\qquad \mathbb{I}_{\{X=a\}},\qquad\mathbb{I}[X=a], \qquad [X=a], \qquad \ldots $$ $\endgroup$ – Sangchul Lee Nov 25 '20 at 18:10
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$\mathbb I$ is an indicator function. It returns $1$ if the predicate inside is true and $0$ if false.

Sometimes this is written $[P]$, the Iverson bracket.

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