### Abstract

In this paper we obtain necessary and sufficient conditions in order that a linear operator, acting in spaces of measurable functions, should admit an integral representation. We give here the fundamental results. Let (T_{i}, μ_{i}) (i=1,2) be spaces of finite measure, and let (T, μ) be the product of these spaces. Let E be an ideal in the space S(T_{1}, μ_{1}) of measurable functions (i.e., from |e_{1}|≤|e_{2}|, e_{1}∈ S (T_{1}, μ_{1}), e_{2}∈E it follows that e_{1}∈E). THEOREM 2. Let U be a linear operator from E into S(T_{2}, μ_{2}). The following statements are equivalent: 1) there exists a μ-measurable kernel K(t,S) such that (Ue)(S)=∫K(t,S) e(t)d μ(t) (e∈E); 2) if 0≤e_{n}≤∈E (n=1,2,...) and e_{n}→0 in measure, then (Ue_{n})(S) →0 μ_{2} a.e. THEOREM 3. Assume that the function F{cyrillic}(t,S) is such that for any e∈E and for s a.e., the μ_{2}-measurable function Y(S)=∫F{cyrillic}(t,S)e(t)d μ_{1}(t) is defined. Then there exists a μ-measurable function K(t,S) such that for any e∈E we have ∫F{cyrillic}(t,S)e(t)d μ_{1}(t)=∫K(t,S)e(t)d μ_{1}(t)μ_{1}a.e.

Original language | English |
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Pages (from-to) | 129-137 |

Number of pages | 9 |

Journal | Journal of Soviet Mathematics |

Volume | 9 |

Issue number | 2 |

DOIs | |

Publication status | Published - 1 Feb 1978 |

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### Scopus subject areas

- Mathematics(all)