Let $(\mu_n)_{n\geqslant 1}$ and $\mu$ be $\sigma$-finite measures on $(\mathbb{R}^d,\mathscr{B}(\mathbb{R}^d))$ such that $\forall n\geqslant 1$, $\mu_n(\mathbb{R}^d)\leqslant 1$ and $\mu(\mathbb{R}^d)\leqslant 1$. Let $\lambda$ denote the Lebesgue measure on $\mathbb{R}^d$. Suppose that $$\forall n\geqslant 1\quad\exists g_n\in\mathbb{L}_1(\lambda)\quad\forall A\in\mathscr{B}(\mathbb{R}^d)\quad\mu_n(A)=\int_Ag_n\,\mathrm{d}\lambda$$ and $$\exists g\in\mathbb{L}_1(\lambda)\quad\forall A\in\mathscr{B}(\mathbb{R}^d)\quad\mu(A)=\int_Ag\,\mathrm{d}\lambda.$$ Is it true that, if $g_n\xrightarrow[n\to+\infty]{}g$ $\lambda$-a.e. and $\displaystyle\int_{\mathbb{R}^d}g_n\,\mathrm{d}\mu_n\xrightarrow[n\to+\infty]{}\int_{\mathbb{R}^d}g\,\mathrm{d}\mu$, then $$\sup_{A\in\mathscr{B}(\mathbb{R}^d)}|\mu_n(A)-\mu(A)|\xrightarrow[n\to+\infty]{}0~?$$ And if it is, how does one establish this result ?


1 Answer 1


No, suppose on $\mathbb R$ we let $g_n = (1/n)\chi_{[0,n]}, n = 1,2,\dots ,$ and define $\mu_n$ accordingly. Set $\mu = 0, g=0.$ Then $g_n \to 0=g$ pointwise everywhere, and

$$\int_{\mathbb R} g_n \, d\mu_n = \int_{\mathbb R} g_n^2 \,d\lambda = \frac{1}{n}\to 0 = \int_{\mathbb R} g\,d\mu .$$

But for every $n,$ $|\mu_n([0,n])- \mu([0,n]) = 1-0.$

  • $\begingroup$ Could you explain what you mean by "define $\mu_n$ accordingly" ? $\endgroup$
    – ADG
    Nov 9, 2017 at 18:53
  • $\begingroup$ The way you have it: $d\mu_n = g_n\,d\lambda.$ $\endgroup$
    – zhw.
    Nov 9, 2017 at 19:20
  • $\begingroup$ Oh, yes, sorry. I overlooked the $\lambda$ in the second integral. $\endgroup$
    – ADG
    Nov 9, 2017 at 19:30

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