The Martin Boundary

We saw in Chapter 1 that if μ is a measure on S, then the equation % MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGObWaaeWaa8aabaWdbiaadIhaaiaawIcacaGLPaaacqGH9aqp % daWdraWdaeaapeGaam4samaabmaapaqaa8qacaWG4bGaaiilaiaadM % haaiaawIcacaGLPaaacaWGKbGaeqiVd02aaeWaa8aabaWdbiaadMha % aiaawIcacaGLPaaaaSWdaeaapeGaam4uaaqab0Gaey4kIipakmaabm % aapaqaa8qacaWG4bGaeyicI4SaamOqaaGaayjkaiaawMcaaiaacYca % aaa!4DDE!$$ h\left( x \right) = \int_S {K\left( {x,y} \right)d\mu \left( y \right)} \left( {x \in B} \right),$$ where K is the Poisson kernel of B, defines a non-negative harmonic function h on B, and that every such function h has a unique representation of this form. In a more general domain Ω non-negative harmonic functions need not have such a representation involving measures on δ∞Ω.